The magnitude of the boat's acceleration is approximately 1.81 m/s², and its direction is approximately 67.4 degrees north of east. We need to calculate the net force acting on the boat and then apply Newton's second law.
To find the magnitude and direction of the boat's acceleration, we need to calculate the net force acting on the boat and then apply Newton's second law.
Given:
Force exerted by wind = 470 N north
Force exerted by water = 190 N east
Mass of the boat = 280 kg
First, we need to determine the net force acting on the boat by adding the vector components of the forces.
Horizontal component: 190 N (east)
Vertical component: 470 N (north)
The net force can be calculated using the Pythagorean theorem:
Net force = √[(190 N)² + (470 N)²]
= √[36100 N² + 220900 N²]
= √(257000 N²)
≈ 507 N
The direction of the net force can be found using trigonometry:
θ = atan(470 N / 190 N)
= atan(2.4737)
≈ 67.4 degrees north of east
Now, we can calculate the acceleration using Newton's second law:
Net force = mass * acceleration
507 N = 280 kg * acceleration
Acceleration = 507 N / 280 kg
≈ 1.81 m/s²
Therefore, the magnitude of the boat's acceleration is approximately 1.81 m/s², and its direction is approximately 67.4 degrees north of east.
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A shot putter releases the shot some distance above the lovel ground with a velooly of 12.0 51.0 " above the horizontal. The shot hits the 51.0 " above the horizontal. The shot hits the ground 2.08 s later. You can ignore air resistance. 7 Part D Whan is the roonponent of the shors velocty at the beginning of as tajectory?
The component of the shot's velocity at the beginning of its trajectory is the horizontal velocity component Vx = 7.83 m/s. Therefore, the answer is 7.83 m/s.
Given information:
A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s at 51.0° above the horizontal.
The shot hits the ground 2.08 s later. We are to find the component of the shot's velocity at the beginning of its trajectory.
Determination of horizontal velocity component
Let the horizontal component of velocity be Vx and the vertical component of velocity be Vy.Let the initial velocity V be 12.0 m/s and the angle of projection θ be 51°.Vx = V cosθ
Here, V = 12.0 m/s, θ = 51°Vx = 12.0 cos 51°= 7.83 m/s
Determination of vertical velocity component
Vy = V sinθ
Here, V = 12.0 m/s, θ = 51°Vy = 12.0 sin 51°= 9.20 m/s
Determination of the component of the shot's velocity at the beginning of its trajectory
The component of the shot's velocity at the beginning of its trajectory is the horizontal velocity component Vx = 7.83 m/s. Therefore, the answer is 7.83 m/s.
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An Atomic Force Microscope (AFM) can produce very detailed surface images ... down to the atomic level. The image, below left, shows the probe, which has a natural frequency of, f
o
=16.2 Hz, which is hooked to a lever arm with a spring constant of, k=1000 N/m, and is driven at f=55 Hz. The resulting amplitude of motion is A=0.42∗10
−10
m. a) From the values above, estimate the mass of the oscillating probe.
m
=9.65∗10
−2
kg b) Take b=0 and estimate the peak force in Newtons.
Fo
=4.42∗10
−7
N
The mass of the oscillating probe is 9.65 × 10⁻² kg and the peak force in Newtons is 4.42 × 10⁻⁷ N.
The mass of the oscillating probe is 9.65 × 10⁻² kg and the peak force in Newtons is 4.42 × 10⁻⁷ N. Here's how to arrive at these answers :Given:
f₀ = 16.2 Hz
k = 1000 N/m
f = 55 Hz
A = 0.42 × 10⁻¹⁰ m
We know that for a simple harmonic motion, amplitude (A) is related to force constant (k), frequency (f) and mass (m) by the following expression:
A = F/mω²where
F = kx (x is displacement from equilibrium position), and ω is the angular frequency.
Rearranging this equation, we get:F = mω²ASubstituting the given values, we get:
F = kAω²/mSubstituting the values of k, A, f and f₀ and taking b = 0 (no damping), we get:
Fo = (1000 × 0.42 × 10⁻¹⁰ × (55/16.2)²)/9.81
= 4.42 × 10⁻⁷ NTo estimate the mass, we need to rearrange the first equation to get:
m = k/f²
A = (1000/16.2²) × 0.42 × 10⁻¹⁰ / (55/16.2)²
= 9.65 × 10⁻² kg
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An object (object A) moves at constant velocity (MEANING?). This object collides with a much smaller non-rigid object (object B) which splats against the larger object. Explain the following from the perspective of Newton's Laws of motion and the Impulse-momentum Theorem. a. Which object experiences the greater force, if at all? b. Which object experiences the greater change in momentum, if at all? c. Which object has a larger impulse transferred to it, if at all? d. Which object experiences the greater change in velocity after the collision? [Think in terms of the magnitudes of the acceleration experienced by the objects.]
The answers to the given questions are as follows:
a. Object B experiences a greater force than object A during the collision.
b. Object B undergoes a greater change in momentum than object A.
c. Object B receives a larger impulse compared to object A.
d. Object B experiences a greater change in velocity after the collision due to its larger acceleration.
a. According to Newton's Laws of Motion, the force experienced by an object is equal to the rate of change of momentum. In this case, since object A is moving at a constant velocity, its momentum remains constant. On the other hand, when object B collides with object A, it experiences a change in momentum due to the collision. Therefore, object B experiences a greater force than object A.
b. The change in momentum of an object can be calculated using the Impulse-momentum Theorem, which states that the change in momentum of an object is equal to the impulse exerted on it. Since object B is smaller and collides with object A, it experiences a larger impulse than object A. As a result, object B undergoes a greater change in momentum compared to object A.
c. The impulse exerted on an object is equal to the product of the force applied to the object and the time interval over which the force acts. In this case, since object B experiences a greater force during the collision, and the time interval is the same for both objects, object B receives a larger impulse compared to object A.
d. The change in velocity of an object can be related to its acceleration using the equation Δv = aΔt, where Δv is the change in velocity, a is the acceleration, and Δt is the time interval. Since object B experiences a greater force and impulse, it undergoes a larger acceleration during the collision. Therefore, object B experiences a greater change in velocity after the collision compared to object A.
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Match the following: Science - Application Engineering - Basic relationship High toughness - Metals High strength - ceramics High specific strength - composites
Science - Application
Engineering - Basic relationship
High toughness - Metals
High strength - ceramics
High specific strength - composites.
Science refers to a body of knowledge that is used to learn about the natural world. It includes a range of subjects like biology, physics, chemistry, geology, astronomy, etc. Science is used to understand the mechanisms and interactions of the natural world and to make predictions about how it will behave in the future. Applications of science can be seen in fields such as medicine, engineering, and technology.
Engineering is the study of design, construction, and use of machines and structures. It includes various fields like mechanical, civil, electrical, chemical, and aerospace engineering. Engineering applies scientific principles to solve practical problems and create new devices, products, and systems. In other words, it takes the knowledge of science and applies it to real-world problems.
High toughness is a property that can be seen in metals. It refers to the ability of a metal to resist fracture or deformation under stress. High strength is a property that can be seen in ceramics. It refers to the ability of a material to resist deformation under stress. Composites, on the other hand, have high specific strength, which means that they are strong in relation to their weight. Composites are materials made up of two or more different types of materials that are combined to create a new material with specific properties.
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What is the potential difference (in Volts) ΔV=V
A
−V
B
between point A, situated 6 cm and point B, situated 16 cm from a 1nC point charge? You should round your answer to an integer, indicate only the number, do not include the unit.
The potential difference between point A and point B is 1.5 volts. The potential difference between two points is the work that is done per unit charge (in volts). For finding the potential difference between point A and point B that are located at a distance of 6 cm and 16 cm from a 1 nC point charge, we need to use the formula: ΔV=V A −V B
The potential difference between two points is the work that is done per unit charge (in volts). For finding the potential difference between point A and point B that are located at a distance of 6 cm and 16 cm from a 1 nC point charge, we need to use the formula: ΔV=V A −V B
where, ΔV is the potential difference in volts.VA is the electric potential at point A.VB is the electric potential at point B. Here, k= 9 × 10 9 Nm 2 C − 2 ,Q = 1 × 10 − 9 C, dA = 6 cm and dB = 16 cm.
From the electric potential formula, we know that: V=kQr
where, k = Coulomb’s constant (9 × 10 9 Nm 2 C − 2 )
Q = Charge of the point in Coulombs
r = Distance from the point in meters
Now, using the formula for potential difference, we can write: ΔV = kQ(1/dA - 1/dB)
ΔV = (9 × 10 9 Nm 2 C − 2 ) × 1 × 10 − 9 C × (1/0.06 m − 1/0.16 m)
ΔV = (9 × 10 9 Nm 2 C − 2 ) × 1 × 10 − 9 C × (16 − 6)/0.06 × 0.16
ΔV = (9 × 10 9 Nm 2 C − 2 ) × 1 × 10 − 9 C × 166.66666666666666
ΔV = 1.5 V
Therefore, the potential difference between point A and point B is 1.5 volts.
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The maximum torque experienced by a coil in a 1.1-T magnetic field is 6.5×10
−4
N⋅m. The coil is circular and consists of only one turn. The current in the coil is 5.3 A. What is the length of the wire from which the coil is made? Number Units
The length of the wire from which the coil is made is 150 m.
Maximum torque, τ = 6.5 × 10⁻⁴ N.m
Magnetic field, B = 1.1 T
Current in the coil, I = 5.3 A
Number of turns, N = 1
Torque, τ = NIBA
Where,L = length of the wire
A = area of the wire`NIB
A = 2πrN * B * I * A``NIB
A = 2πrN * B * I * πr²``NIB
A = 2πr³ * B * I``r³ = τ / (2πNI * B)`
Length of wire, L = 2πr
Length, `L = 2π * (τ / 2πNI * B) = τ / (NI * B)`
Putting the values,L = 6.5 × 10⁻⁴ / (1 * 5.3 * 1.1)
L = 150 m
Hence, the length of the wire from which the coil is made is 150 m.
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uniform board of length 14 m weighing 250 N is supported by ropes at each end. A boy weighing 600 N stands 5.0 m from the rope on the left. Determine the tension in the two ropes for static equilibrium.
Therefore, the tension in the left rope (T1) is 635.71 N and the tension in the right rope (T2) is 214.29 N.
In order to find the tension in the two ropes for static equilibrium, we need to determine the forces acting on the uniform board and apply the conditions of equilibrium.
Firstly, let's draw a free-body diagram of the board.
Free body diagram of the board
Here, T1 and T2 are the tensions in the ropes at the left and right ends respectively, W1 is the weight of the board, and W2 is the weight of the boy.
Now, let's apply the conditions of equilibrium.
The conditions of equilibrium are:
1. The net force acting on the object must be zero.
2. The net torque acting on the object about any point must be zero.
Taking the sum of the forces in the vertical direction, we get:
T1 + T2 - W1 - W2 = 0
Substituting the values we get:
T1 + T2 - 250 - 600 = 0
T1 + T2 = 850 N
Taking moments about point A, we get:
T2 × 14 - 600 × 5 = 0
T2 = (600 × 5) / 14
T2 = 214.29 N
Substituting the value of T2 in the first equation, we get:
T1 + 214.29 = 850
T1 = 635.71 N
Therefore, the tension in the left rope (T1) is 635.71 N and the tension in the right rope (T2) is 214.29 N.
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When a ton of coal burns, it loses 3.3 x 10^-7 kg of mass. The
energy given off is 3.0 x 10^10 J. Why?
The reason why a ton of coal loses 3.3 x 10^-7 kg of mass when it burns is because of mass-energy equivalence. This is a principle of physics that states that mass and energy are equivalent, and that they can be converted into each other.
The equation that represents this principle is E = mc^2, where E is energy, m is mass, and c is the speed of light. In the case of a ton of coal burning, the energy given off is 3.0 x 10^10 J. This energy is equivalent to a mass of 3.3 x 10^-7 kg, according to the equation E = mc^2. So, when the coal burns, some of its mass is converted into energy. The mass that is converted into energy is very small, but it is measurable. This is why scientists are able to measure the mass-energy equivalence of different substances. The mass-energy equivalence principle was first proposed by Albert Einstein in 1905. It has since been verified by many experiments, and it is now one of the fundamental principles of physics.
The mass-energy equivalence principle has many important implications not separate entities, but are instead two different forms of the same thing.
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A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.05 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 73.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j^ direction is upward.) Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. j^ N
The normal force exerted by the floor of the elevator on the student during her brief acceleration is 716.35 N in the upward direction (j^).
the normal force exerted by the floor of the elevator on the student during the upward acceleration, we need to consider the forces acting on the student.
The forces involved are the force of gravity (mg) and the normal force (N). Since the elevator and the student are accelerating upwards, there is an additional upward force, ma, due to the acceleration.
Using Newton's second law, we can write the equation of motion in the vertical direction:
ΣFy = N - mg - ma = 0
Rearranging the equation:
N = mg + ma
Substituting the given values:
m = 73.0 kg
g = 9.8 m/s² (acceleration due to gravity)
a = 5.05 m/s² (upward acceleration)
N = (73.0 kg) * (9.8 m/s²) + (73.0 kg) * (5.05 m/s²)
Calculating the expression:
N ≈ 716.35 N
The normal force exerted by the floor of the elevator on the student during the brief upward acceleration is 716.35 N.
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Show that harmonic wavestudent submitted image, transcription available below(x,t) =A sin k(x-vt) is a solution of the 1D differential wave equation.
The function (x, t) = A sin(k(x - vt)) is a solution of the one-dimensional differential wave equation as it satisfies the equation when substituted into it. It demonstrates that the function describes harmonic waves.
To show that the function (x, t) = A sin(k(x - vt)) is a solution of the one-dimensional differential wave equation, we need to substitute it into the wave equation and verify that it satisfies the equation.
The one-dimensional wave equation is given by:
∂²/(∂x²) - (1/v²) ∂²/(∂t²) = 0,
where ∂²/(∂x²) represents the second partial derivative with respect to x, and ∂²/(∂t²) represents the second partial derivative with respect to t.
Let's begin by calculating the first and second derivatives of (x, t) with respect to x and t:
∂/(∂x) (x, t) = A k cos(k(x - vt)),
∂/(∂t) (x, t) = -A v k cos(k(x - vt)),
∂²/(∂x²) (x, t) = -A k² sin(k(x - vt)),
∂²/(∂t²) (x, t) = A v² k² sin(k(x - vt)).
Now, substitute these derivatives into the wave equation:
(-A k² sin(k(x - vt))) - (1/v²) (A v² k² sin(k(x - vt))) = 0.
Simplifying the equation:
-A k² sin(k(x - vt)) - A k² sin(k(x - vt)) = 0.
Since the two terms on the left-hand side are equal, the equation holds true.
Therefore, we have shown that (x, t) = A sin(k(x - vt)) satisfies the one-dimensional differential wave equation, confirming it as a solution.
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A compounded reverted gear train is to be designed as a step up gear to increase the speed by 48 times. With pressure angle of 20
∘
, specify suitable numbers of teeth to minimize gearbox size and avoid interference problem in the teeth. Sketch the designed compounded reverted gear system, indicating the gear positions and their numbers of teeth.
The first gear set (12-72) provides a speed increase of 6, and the second gear set (8-64) further increases the speed by a factor of 8, resulting in an overall speed increase of 48.
To design a compounded reverted gear train as a step-up gear with a speed increase of 48 times while minimizing the gearbox size and avoiding interference problems, we need to carefully select the numbers of teeth for the gears.
The compounded reverted gear train consists of two sets of gears: the first set is the compound gear train, and the second set is the reverted gear train. The compound gear train is used to achieve a moderate speed increase, and the reverted gear train further amplifies the speed.
To determine suitable numbers of teeth, we can start by considering the speed increase ratio of 48. This ratio can be achieved by breaking it down into smaller ratios for the compound and reverted gear sets. For example, we can choose a speed increase ratio of 6 in the compound gear train and a ratio of 8 in the reverted gear train.
Next, we need to select suitable numbers of teeth for each gear to avoid interference problems and ensure smooth operation. To minimize gearbox size, we aim to keep the gears as small as possible. One approach is to use gears with a small number of teeth, which can help reduce their size.
For the compound gear train, we can select gears with, for example, 12 and 72 teeth. This gives a speed increase ratio of 6. For the reverted gear train, we can choose gears with 8 and 64 teeth, resulting in a speed increase ratio of 8.
The sketch of the designed compounded reverted gear system would show the gear positions and their numbers of teeth as follows:
---- 12 teeth ----
/ \
--- 72 teeth 8 teeth ---
\ /
---- 64 teeth ----
In this configuration, the first gear set (12-72) provides a speed increase of 6, and the second gear set (8-64) further increases the speed by a factor of 8, resulting in an overall speed increase of 48. By carefully selecting the numbers of teeth and the gear ratios, we can achieve the desired speed increase while minimizing the size of the gearbox and avoiding interference problems between the gear teeth.
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A 13 g coin slides upward on a surface that is inclined at an angle of 27 ∘ above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.22; the coefficient of static friction is 0.24.
A. Find the magnitude of the force of friction when the coin is sliding upward initially.
B. Find the magnitude of the force of friction when the coin is sliding back downward later.
A. Initially, when the coin is sliding upward, the magnitude of the force of friction can be determined using the equation:
Frictional force = μk * Normal force,
where μk is the coefficient of kinetic friction and Normal force is the perpendicular force exerted by the surface on the coin. To calculate the Normal force, we need to consider the forces acting on the coin. The weight of the coin can be decomposed into two components:
one parallel to the inclined surface (mg * sinθ) and one perpendicular to it (mg * cosθ), where m is the mass of the coin and θ is the angle of inclination. Since the coin is sliding, the normal force is equal in magnitude and opposite in direction to the perpendicular component of the weight.
Using the given values:
Mass of the coin (m) = 13 g = 0.013 kg
Angle of inclination (θ) = 27°
Coefficient of kinetic friction (μk) = 0.22
The Normal force = mg * cosθ = 0.013 kg * 9.8 m/s^2 * cos(27°)
Finally, the magnitude of the force of friction can be calculated as:
Frictional force = μk * Normal force.
B. When the coin is sliding back downward, the magnitude of the force of friction can be determined using the same equation. However, in this case, we need to consider the coefficient of static friction (μs) since the coin is at the point of impending motion.
The coefficient of static friction (μs) is greater than the coefficient of kinetic friction (μk). Therefore, the magnitude of the force of friction when the coin is sliding back downward will be higher compared to the magnitude calculated in part A.
However, the exact value cannot be determined without knowing the angle of inclination, as it affects the decomposition of the weight and the calculation of the Normal force. when the coin is initially sliding upward, the magnitude of the force of friction can be calculated using the coefficient of kinetic friction and the Normal force.
When the coin is sliding back downward, the magnitude of the force of friction will be higher, but the exact value depends on the angle of inclination, which is not provided.
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1. The only accurate method of charging is to;
A. charge until the sight glass is clear
B. charge until gauge pressures are correct
C. weigh the refrigerant into the system
D. charge as a vapor with the engine running
2. Which of the following refrigerant has the lowest GWP?
A. R410A
B. R134A
C. R152a
D. R744
The only accurate method of charging is to weigh the refrigerant into the system. The correct option is c.
R152a has the lowest GWP (Global Warming Potential). The correct option is c.
1. The only accurate method of charging is to weigh the refrigerant into the system. While installing a new or retrofit air conditioning system, refrigerant is the crucial component that must be handled with care. Overcharging or undercharging will cause the system to underperform or, in the worst-case scenario, malfunction.
When charging the refrigerant, the only accurate method is to weigh it into the system using an accurate refrigerant scale that can measure the correct amount to within a tenth of an ounce.
2. R152a has the lowest GWP (Global Warming Potential) of the refrigerants listed. It is used as a refrigerant in various applications, including domestic and automotive air conditioning systems. R152a is a hydrofluorocarbon that has an insignificant impact on the environment and a global warming potential of just 124.
It is a non-ozone-depleting compound that has no impact on the ozone layer. It's a much better choice than some of the earlier-generation refrigerants.
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The most accurate method of charging a refrigeration system is to weigh the refrigerant into the system. The refrigerant with the lowest global warming potential is R744, also known as Carbon Dioxide.
Explanation:The only accurate method of charging a refrigeration system is to weigh the refrigerant into the system (option C). This is because different refrigerants and systems have specific weight requirements for optimal performance. Relying solely on the sight glass or gauge pressure may lead to overcharging or undercharging, which can impact system effectiveness and efficiency.
Regarding the refrigerant with the lowest Global Warming Potential (GWP), it would be R744 (option D), also known as Carbon Dioxide (CO2). Its GWP is essentially 1, substantially lower than the other refrigerants listed. Note that lower GWP is more environmentally friendly, contributing less to global warming.
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A gymnast with a mass of 72.5 kg is climbing a rope.
a. (1 pt.) Draw a free body diagram of the gymnast when they’re climbing at
constant speed.
b. (1 pt.) Find the tension that the gymnast puts on the rope when they’re
climbing at constant speed
c. (1 pt.) Draw a free body diagram of the gymnast when they’re accelerating
upwards at a rate of 1.63 m/s2.
d. (1 pt.) What is the tension on the rope now?
a. The free body diagram shows forces of gravity (mg), tension in the rope (T), and the normal force (N).
b. The tension in the rope when climbing at constant speed is equal to the force of gravity, T = mg.
c. When accelerating upwards, the free body diagram includes the same forces as before with an additional force representing acceleration.
d. The tension on the rope when accelerating upwards is found by T = ma + mg, using the given values.
a. When the gymnast is climbing at constant speed, the free body diagram would show the following forces: the force of gravity acting downward (mg), the tension force in the rope acting upward (T), and the normal force (N) exerted by the rope on the gymnast.
b. When the gymnast is climbing at constant speed, the tension in the rope is equal to the force of gravity acting on the gymnast. Therefore, the tension is given by T = mg, where m is the mass of the gymnast and g is the acceleration due to gravity.
c. When the gymnast is accelerating upwards at a rate of 1.63 m/s², the free body diagram would include the same forces as before (mg, T, N), but now an additional force in the upward direction representing the acceleration (ma).
d. To determine the tension on the rope when the gymnast is accelerating upwards, we need to consider the net force acting on the gymnast. The net force is given by the equation: ΣF = ma, where m is the mass of the gymnast and a is the acceleration. The tension force in the rope can be found by rearranging the equation: T = ma + mg. Substitute the given values to calculate the tension.
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A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 5.9 A. Determine the power consumed by (a) the blow-dryer and (b) the vacuum cleaner. (c) Determine the ratio of the energy used by the blow-dryer in 19 minutes to the energy used by the vacuum cleaner in 48 minutes. (a) Number Units (b) Number Units (c) Number Units
The required ratio of energy used by the blow-dryer in 19 minutes to the energy used by the vacuum cleaner in 48 minutes is 0.805.
Given values: Voltage, V = 120 V
Current, I (blow-dryer) = 12 A
Current, I (vacuum cleaner) = 5.9 A
(a) Power consumed by blow-dryer:Formula: P = VI
Where P is the power, V is the voltage, and I is the current.Substituting the given values:
P = 120 V x 12 AP = 1440 W
The power consumed by the blow-dryer is 1440 W.
(b) Power consumed by vacuum cleaner:Formula: P = VI
Where P is the power, V is the voltage, and I is the current.Substituting the given values:P = 120 V x 5.9 AP
= 708 W
The power consumed by the vacuum cleaner is 708 W.
(c) Ratio of energy used by blow-dryer in 19 minutes to energy used by vacuum cleaner in 48 minutes:Formula: Energy consumed = Power x time
Where energy consumed is in watt-hours (Wh), power is in watts (W), and time is in hours.Substituting the given values,Let E1 be the energy consumed by blow-dryer.
E1 = 1440 W x (19/60) hours
E1 = 456 Wh
Let E2 be the energy consumed by the vacuum cleaner.
E2 = 708 W x (48/60) hours
E2 = 566.4 Wh
Ratio of energy consumed = E1 / E2
Ratio of energy consumed = 456 Wh / 566.4 Wh
Ratio of energy consumed = 0.805
The required ratio of energy used by the blow-dryer in 19 minutes to the energy used by the vacuum cleaner in 48 minutes is 0.805.
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(C) A particle is projected at an angle of 30∘ above the horizontal with a speed of 50 ms^−1 . Determine (i) the height at horizontal point of 2 m. (4 marks) (ii) range of the projectile (4 marks) (iii) the maximum height reached. (4 marks)
When an object is projected at an angle, the motion is divided into two parts: one is the vertical motion, and the other is the horizontal motion. Here, the object is projected at an angle of 30° above the horizontal with a speed of 50 m/s, as shown below:
The initial velocity of the particle is
v0 = 50 m/s at 30°
(i) Height at the Horizontal Point of 2 m:
To calculate the height at the horizontal point of 2 m, we can use the formula for vertical motion:
[tex]y = yo + vot + 1/2 gt^2[/tex]
where y is the vertical displacement, yo is the initial vertical position, vo is the initial velocity, t is the time, and g is the acceleration due to gravity. At the horizontal point, the vertical displacement of the particle is 2 m, the initial vertical position is zero, and the initial velocity is v0 sinθ.
Hence, we can write the above equation as:
2 = 0 + v0 sinθt + 1/2 [tex]gt^2[/tex]
t = 2v0 sinθ/g = 2(50)(sin 30°)/(9.8) = 5.11 s
Now, we can calculate the height at the horizontal point using the above equation:
[tex]y = yo + vot + 1/2 gt^2[/tex]
y = 0 + v0 sinθt + 1/2 [tex]gt^2[/tex]
[tex]y = (50)(sin 30°)(5.11) - 1/2 (9.8)(5.11)^2[/tex]
y = 78.48 m
Therefore, the height at the horizontal point of 2 m is 78.48 m.
(ii) Range of the Projectile:
To calculate the range of the projectile, we can use the formula for horizontal motion:
x = xo + v0x t
where x is the horizontal displacement, xo is the initial horizontal position, v0x is the initial horizontal velocity, and t is the time. At the highest point of the particle, the horizontal displacement is the range of the projectile, the initial horizontal position is zero, and the initial horizontal velocity is v0 cosθ.
Hence, we can write the above equation as:
R = 0 + v0 cosθ t
t = R/v0 cosθ
Substituting the value of t in the equation for vertical motion:
[tex]y = 0 + v0 sinθ (R/v0 cosθ) - 1/2 g (R/v0 cosθ)^2[/tex]
[tex]R = v0^2 sin 2θ/g[/tex]
[tex]R = (50)^2 sin 60°/9.8[/tex]
R = 255.1 m
Therefore, the range of the projectile is 255.1 m.
(iii) Maximum Height Reached:
To calculate the maximum height reached by the particle, we can use the formula for vertical motion:
[tex]y = yo + vot + 1/2 gt^2[/tex]
At the highest point, the vertical velocity of the particle becomes zero. Therefore, we have:
0 = v0 sinθ + gt
t = v0 sinθ/g
Substituting the value of t in the equation for vertical motion:
[tex]y = 0 + v0 sinθ (v0 sinθ/g) - 1/2 g (v0 sinθ/g)^2[/tex]
[tex]y = v0^2 sin^2 θ/2g[/tex]
[tex]y = (50)^2 (sin 30°)^2/2(9.8[/tex]
y = 63.7 m
Therefore, the maximum height reached by the particle is 63.7 m.
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If a current of 1.5 A flows in a constantan wire of length 1.0 m and cross sectional area of 0.50 mm
2
, determine the potential difference across the two ends of the wire. Resistivity of constantan is 4.9×10
−7
Ωm. (1.47 V) 2. The resistance of a coil of copper wire is 3.35Ω at 0
∘
C. What is the resistance at 50
∘
C. For copper, α=4.3×10
−3
∘
C
−1
⋅(4.1Ω) 3. A cell has an emf of 1.54 V. when it is in series with a 1.0Ω resistance, the reading of a voltmeter connected across the cell terminals is 1.40 V. Determine the cell's internal resistance. (0.10ΩΩ) 4. A dry cell delivering 1.5 A current has a terminal potential difference of 1.25 V. What is the internal resistance of the cell if its open circuit potential (a.k.a. emf) is 1.50 V ? 5. An electric heater of 5.0Ω is immersed totally in a beaker containing 0.50 kg of water at 25
∘
C. A 24 V dc supply is connected to the heater for 5 minutes. a) Determine the electrical power generated by the heater b) Determine the current in the heater (115 W, 4.8 A) 6. The copper wire used in a house has a cross-sectional area of 3.00 mm
2
. If 10.0 m of this wire is used to wire a circuit in the house at 20.0
∘
C, find the resistance of the wire at temperatures of (a) 30.0
∘
C and (b) 10.0
∘
C. ( rho
copper
=1.7×10
−8
Ω.m,α
copper
=3.9×10
−3
) (0.059Ω,0.055Ω ) 7. The cost of electricity varies widely throughout the United States; $0.120/kWh is a typical value. At this unit price, calculate the cost of (a) leaving a 40.0- W porch light on for 2 weeks while you are on vacation, (b) making a piece of dark toast in 3.00 min with a 970−W toaster, and (c) drying a load of clothes in 40.0 min in a 5 200-W dryer. (\$1.61, \$0.00582, \$0.416) 8. A particular wire has a resistivity of 3.0×10
−8
Ω⋅m and a cross-sectional area of 4.0×10
−6
m
2
. A length of this wire is to be used as a resistor that will develop 48 W of power when connected across a 20−V battery. What length of wire is required? (1.11 km)
1) The potential difference across the two ends of the wire is 14.7 V. 2) the resistance of the copper coil at 50 °C is approximately 4.1 Ω.
1) Ohm's Law states that the potential difference (V) across a resistor is equal to the product of its resistance (R) and the current (I) flowing through it: V = I × R.
The current flowing through the wire is 1.5 A and the length of the wire is 1.0 m, we can calculate the resistance using the formula: R = resistivity × (length / cross-sectional area).
Resistivity is given as 4.9 × 10^(-7) Ωm, the length is 1.0 m, and the cross-sectional area is 0.50 mm^2, which is equivalent to 5.0 × 10^(-7) m^2.
R = (4.9 × 10^(-7) Ωm) × (1.0 m / 5.0 × 10^(-7) m^2)
R = 9.8 Ω
Now, the potential difference (V) across the wire using Ohm's Law:
V = (1.5 A) × (9.8 Ω)
V = 14.7 V
Therefore, the potential difference across the two ends of the wire is 14.7 V.
2. The resistance of the copper coil at 50 °C, we can use the formula:
Rt = R0 × (1 + α × ΔT)
where Rt is the resistance at temperature t, R0 is the resistance at the reference temperature (0 °C in this case), α is the temperature coefficient of resistance for copper, and ΔT is the temperature difference in °C.
The resistance of the copper coil at 0 °C is 3.35 Ω and the temperature coefficient of resistance for copper (α) is 4.3 × 10^(-3) °C^(-1), we can calculate the resistance at 50 °C:
R50 = 3.35 Ω × (1 + 4.3 × 10^(-3) °C^(-1) × (50 °C - 0 °C))
Simplifying the equation gives:
R50 ≈ 4.1 Ω
Therefore, the resistance of the copper coil at 50 °C is approximately 4.1 Ω.
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In the experiment of linear kinematics, the materials are given: air track, cart, photo gates, calculator, graph paper. Give 3 sources of error in linear kinematics experiment, and give detailed explanation (Calculation errors or rounding do not count).
Sources of error in the linear kinematics experiment include environmental factors (such as air resistance, temperature, and friction), human error in data recording and time measurements, and the potential for experimenter bias in result interpretation. These factors can introduce inaccuracies and should be accounted for to ensure reliable experimental outcomes.
In the experiment of linear kinematics, some sources of error include the following:
Environmental Factors: There are numerous environmental factors that may influence the outcomes of the experiment. These include factors like air resistance, the temperature of the lab, and friction between the air track and the cart. These may cause the cart to slow down at a faster rate than it should, resulting in an incorrect measurement of the acceleration.Human Error: Human error is another significant source of error. This might happen when recording data, where there may be deviations in the measurements. This type of error might also happen when using the photo gates to determine the time interval, which could lead to a deviation in measurement.Experimenter Bias: Experimenter bias is when the person running the experiment allows their thoughts to influence the experiment's results. It may occur in several ways, but the most frequent is that the experimenter may select the outcomes that better align with their theory. They might subconsciously dismiss or ignore findings that do not fit their hypothesis.Learn more about air resistance: https://brainly.com/question/9143942
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Write a reply post of venus and its acceleration of gravity. Assume that all planets are habitable like the Earth, and they have a solid surface. Assume that we are on Venus. The acceleration of gravity is different from the acceleration on the Earth. What would you find easier to do or more challenging to do on Venus with respect to when you are on Earth?
The acceleration of gravity on Venus is almost 91% of Earth's acceleration of gravity. It would be easier to carry things than on Earth, due to its thick atmosphere, walking, and running would be more challenging on Venus
Venus and its acceleration of gravity: In terms of acceleration of gravity, Venus is the second-most considerable planet in the solar system, with a gravitational pull approximately 0.904 times that of Earth's acceleration of gravity. On Venus, a 100-pound object would weigh 90.4 pounds (41 kilograms).
Venus is a rocky planet with a surface that is roughly 90% of Earth's mass and 81.5% of Earth's radius. Its dense, cloud-covered atmosphere, which is primarily composed of carbon dioxide, produces a surface temperature of nearly 864 degrees Fahrenheit (462 degrees Celsius).
Venus is the most inhospitable planet in the solar system because of its severe climate and the toxic atmosphere it possesses.
The acceleration of gravity on Venus is almost 91% of Earth's acceleration of gravity. The formula for acceleration of gravity is g = GM/r2, where G is the gravitational constant, M is the mass of the object, and r is the distance between the two objects' centers of mass.
G has a constant value, while the mass and radius of Venus are constant. As a result, the acceleration of gravity is proportional to the mass of the planet divided by its radius squared (GM/r2).
The acceleration of gravity on Venus is less than that on Earth, implying that if an individual weighs 100 pounds on Earth, they would weigh 90.4 pounds on Venus.
This means that on Venus, it would be easier to carry things than on Earth. As a result, working with heavy items would be easier on Venus than on Earth.
Due to its thick atmosphere, walking, and running would be more challenging on Venus, and it would take more effort to do so. Running or walking on Venus would feel like running or walking uphill on Earth, making it more difficult to move around. As a result, walking or running on Venus is more challenging than on Earth.
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What are the magnitude and direction of an electric field that exerts a 3.75×10
−5
N upward force on a −2.45μC charge? magnitude N/C direction
To determine the magnitude and direction of the electric field, we can use the equation:
Electric Field (E) = Force (F) / Charge (q)
The given force is 3.75×10^(-5) N, and the charge is -2.45μC, which can be written as -2.45×10^(-6) C.
Substituting these values into the equation, we get:
Electric Field (E) = (3.75×10^(-5) N) / (-2.45×10^(-6) C)
Calculating this expression gives us:
E ≈ -15.31 N/C
The magnitude of the electric field is approximately 15.31 N/C. Note that the negative sign indicates that the electric field is directed in the opposite direction of the force.
Since the force is upward, the electric field is directed downward.
Therefore, the magnitude of the electric field is 15.31 N/C, and its direction is downward.
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A 2.7 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determine the force constant (in N/m ) of the spring, if the box compresses the spring 5.5 cm before coming to rest. N/m (b) Determine the initial speed (in m/s ) the box would need in order to compress the spring by 1.8 cm.
To determine the force constant of the spring, we can use Hooke's Law. The force constant of the spring is approximately 4827.27 N/m. The initial speed required to compress the spring by 1.8 cm is approximately 1.07 m/s.
(a) To determine the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is given by:
F = -kx
where F is the force exerted by the spring, k is the force constant of the spring, and x is the displacement of the spring.
In this case, when the box compresses the spring by 5.5 cm (which is equivalent to 0.055 m), it comes to rest. At this point, the force exerted by the spring is equal to the weight of the box, which can be calculated as:
F = mg
where m is the mass of the box and g is the acceleration due to gravity.
Therefore, we can set up the equation:
mg = -kx
Solving for k, we get:
k = -mg/x
Substituting the given values, we have:
k = -(2.7 kg)(9.8 m/s^2)/(0.055 m)
k ≈ -4827.27 N/m
Note that the negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.
Therefore, the force constant of the spring is approximately 4827.27 N/m.
(b) To determine the initial speed required to compress the spring by 1.8 cm (which is equivalent to 0.018 m), we can use the same formula as before:
F = -kx
At the maximum compression, when the box momentarily comes to rest, the force exerted by the spring is equal to the weight of the box:
mg = -kx
Solving for v (the initial speed), we can use the principle of conservation of mechanical energy. The initial potential energy of the box when it is at rest is converted into the elastic potential energy stored in the spring when it is compressed.
Therefore, we can set up the equation:
(1/2)mv^2 = (1/2)kx^2
Simplifying and solving for v, we get:
v = sqrt((k/m)x^2)
Substituting the given values, we have:
v = sqrt((4827.27 N/m) / (2.7 kg) * (0.018 m)^2)
v ≈ 1.07 m/s
Therefore, the initial speed required to compress the spring by 1.8 cm is approximately 1.07 m/s.
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A mass is attached to a spring and displaced and then released from rest. Determine the time when the kinetic and potential energies are first equal.
the time when the kinetic and potential energies are first equal is [tex]$\frac{\pi}{2} \sqrt{\frac{m}{k}}$.[/tex]
When a mass attached to a spring is displaced and released from rest, the energy of the system is not conserved since energy is converted from potential energy to kinetic energy and back. The initial potential energy is [tex]$PE = \frac{1}{2} k x_0^2$[/tex]while the initial kinetic energy is [tex]$KE = 0$[/tex]
when the mass is released from rest. Since energy is conserved $PE_i = KE_i + PE_f$ when potential energy is equal to kinetic energy: [tex]$$\frac{1}{2} k x_0^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x^2$$[/tex]
where v is the velocity of the mass and x is the displacement of the mass from its equilibrium position. The maximum displacement of the mass is x0.
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A baseball is thrown straight upward on the Moon (take g=1.60 m/s
2
) with an initial speed of 35 m/s. Compute: A) The maximum height reached by the ball Typeyour answer... 1 point [From Problem Above] B) The time taken to reach that height Type your answer... 1 point [From Problem Above] C) Its velocity 30 s after it is thrown Type your answer... 1 point [From Problem Above] D) Time when the ball height is 100 m ascending
A)
Use the equation v = u + gt,
where v is the final velocity (0 at maximum height), u is the initial velocity (35 m/s), g is the acceleration due to gravity on the moon (1.60 m/s²), and t is the time taken to reach the maximum height.
Set v = 0 and solve for t: 0 = 35 - 1.60t.
Solve the equation to find t: t = 21.875 s.
Substitute the value of t into the equation H = ut - (1/2)gt², where H is the maximum height.
Calculate the maximum height: H = (35 m/s)(21.875 s) - (1/2)(1.60 m/s²)(21.875 s)² = 307.42 m.
B)
The time taken to reach the maximum height was already calculated in step A as t = 21.875 s.
C)
Use the equation v = u + gt, where v is the final velocity, u is the initial velocity (35 m/s), g is the acceleration due to gravity on the moon (1.60 m/s²), and t is the time (30 s) after the ball is thrown.
Calculate the velocity: v = 35 m/s - (1.60 m/s²)(30 s) = -17 m/s.
The negative sign indicates that the ball is moving downward.
D)
Use the equation H = ut - (1/2)gt²,
where H is the height (100 m), u is the initial velocity (35 m/s), g is the acceleration due to gravity on the moon (1.60 m/s²), and t is the time.
Rearrange the equation to form a quadratic equation: 2t² - 70t + 400 = 0.
Solve the quadratic equation to find the value of t: t = 8.438 s.
Therefore, the ball height is 100 m at 8.438 s.
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A cement block accidentally falls from rest from the ledge of a 82.1-m-high building. When the block is 13.8 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
The time for the man to get out of the way, when a cement block accidentally falls from rest from the ledge of a 82.1-m-high building is 2.98 seconds.
Given:
Height of the building, h = 82.1 m
Height of the man, h₂ = 2.00 m
Height of the block above the ground,
y = h - h₂ - 13.8 m = 82.1 - 2 - 13.8 = 66.3 m
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time taken to reach the ground, t = ?
We can use the following kinematic equation to find the time:
`y = ut + (1/2)at²`
We get,`t = √(2y/a)`
Putting in the values,`
t = √(2×66.3/9.81)
`t = 3.24 s
But this is the time for the block to reach the ground. The man has to move out of the way before the block reaches him. Hence, he has less time. We can use the following kinematic equation to find the time taken by the block to cover the last 13.8 m:
`y = ut + (1/2)at²`
Here,`y = 13.8 m`, `u = 0 m/s`, `a = 9.81 m/s²`.
We get,
`t = √(2y/a)`
Putting in the values,
`t = √(2×13.8/9.81)
`t = 1.21 s
Therefore, the man has at most 1.21 seconds to move out of the way before the cement block hits him. So, he has 3.24 - 1.21 = 2.03 seconds to move out of the way.
When a cement block accidentally falls from rest from the ledge of a 82.1-m-high building, the man, 2.00 m tall, who is 13.8 m away from the cement block would have at most 1.21 seconds to move out of the way.
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Siga enters a straight- air path at the 2.9km mark and flies west with a constant velocity of 5.1m/s. AT THE SAME TIME John flies east from the .2km mark of the same path at a velocity of 6m/s. The path runs from west (lower numbers) to east (higher numbers).
a.) Write a function x(t) that describes the position of Siga as a function of time with respect to the straight-air path.
b.) Write a function x(t) that describes the position of John as a function of time with respect to the straight-air path.
c.) At what position on the path do they meet?
Therefore, they meet at a position on the path that is 1463.854 m from the 2.9 km mark where Siga entered the path.
Given,Siga enters a straight-air path at the 2.9 km mark and flies west with a constant velocity of 5.1 m/s.John flies east from the 0.2 km mark of the same path at a velocity of 6 m/s.The function x(t) that describes the position of Siga as a function of time with respect to the straight-air path is:
x(t) = 2900 - 5.1t
Here, 2900 is the initial position of Siga when he enters the path at the 2.9 km mark.
The function x(t) that describes the position of John as a function of time with respect to the straight-air path is:x(t) = 200 + 6tHere, 200 is the initial position of John when he enters the path at the 0.2 km mark.They meet at a position on the path when the position of Siga is equal to the position of John.
x(t) = x(t)2900 - 5.1
t = 200 + 6t
Solving for t:2900 = 11.1tt
= 261.26 seconds
Substituting t in the equation of Siga's position:
x(t) = 2900 - 5.1
t= 2900 - 5.1 × 261.26
= 1463.854 m
Therefore, they meet at a position on the path that is 1463.854 m from the 2.9 km mark where Siga entered the path.
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Select the following arrangement of charges that will Attract one another. A positive and a negative charge Two positive charges Two negative charges A positive charge and an uncharged object A negative charge and an uncharged object. Two uncharged objects.
The arrangement of a positive and a negative charge is the only one that will attract each other.
In an electric field, opposite charges attract each other. Positive charges have a tendency to move towards negative charges, and negative charges have a tendency to move towards positive charges.
The other options listed do not result in an attractive force between charges:
Two positive charges will repel each other because they have the same charge.
Two negative charges will repel each other because they have the same charge.
A positive charge and an uncharged object will not experience any significant attractive force.
A negative charge and an uncharged object will not experience any significant attractive force.
Two uncharged objects will not experience any significant attractive force.
Therefore, the arrangement of a positive and a negative charge is the only one that will attract each other.
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Lab Report: Errors and Significant Figures 1. Which one of the measurements (length, depth or width) for the volume of the block of wood was the most precise? Explain. (Hint: which measurement has the smallest percentage of uncertainty?) 2. Which of the volume measurements (block, cylinder, etc.) was more accurate? How could you tell? 3. Propagation of errors: Compute the following value along with its uncertainty. y=x−z Where, x=(100±0.01)mm and z=(50±0.02)mm 4. Propagation of errors: Compute the following value along with its uncertainty. y=
z
x
2
Where, x=(25±0.2)m and z=(125±0.2)m 5. Propagation of errors: Compute the following value along with its uncertainty. a=(x+y)⋅z Where, x=(16.5±0.02)mm,y=(9.3±0.05)mm and z=(42.0±0.2) mm
1. The measurement of the depth of the block is the most precise because it has the smallest percentage of uncertainty. The precision of a measurement is determined by the uncertainty in the measurement.
The smaller the uncertainty, the more precise the measurement. The percentage uncertainty is calculated by dividing the uncertainty by the measurement value and then multiplying it by 100. The percentage uncertainty is the smallest for the depth measurement, indicating that it is the most precise measurement.2. The measurement of the block's volume is more accurate. The cylinder's volume measurement could be compared to the volume measurement of the block of wood to see which was more accurate. The more accurate measurement is the one that is closer to the accepted value. The measurement that is closer to the accepted value is the more accurate measurement.3. y = x - z = 100 - 50 = 50 mmThe uncertainty in y is the sum of the uncertainties in x and z.δy = δx + δz = 0.01 + 0.02 = 0.03 mm
Therefore, y = (50 ± 0.03) mm.4. y = z/(2x) = 125/(2 × 25) = 2.5 m^-1The uncertainty in y is calculated using the following equation:δy/y = δx/x + δz/z = 0.2/25 + 0.2/125 = 0.008Therefore, δy = y × δy/y = 2.5 × 0.008 = 0.02 m^-15. a = (x + y)z = (16.5 + 9.3) × 42 = 1029.6 mm^2The uncertainty in a is calculated using the following equation:δa/a = δx/x + δy/y + δz/z = 0.02/16.5 + 0.05/9.3 + 0.2/42 = 0.022
Therefore, δa = a × δa/a = 1029.6 × 0.022 = 22.6 mm^2Therefore, a = (1029.6 ± 22.6) mm^2.
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A Snake's Journey
A snake travels 8 meters north, 4 meters east, 2 meters west, and 3 megers south. It takes the snake 5 minutes to make the trip.
What is:
1. The snake's total distance traveled?
2. The snake's final displacement?
3. The snake's average speed (answer in m/s)?
4. The snake's average velocity (answer in m/s)?
If your answer is a vector that requires a direction, be sure to include it as well. Unless instructed, leave your answer in standard form.
The snake traveled a total of 17 meters,the snake's final displacement was approximately 5.39 meters,the snake's average speed was approximately 0.057 m/s,
the snake's average velocity was approximately (0.0067 m/s, 0.017 m/s) in the east-north direction.
1. To find the snake's total distance traveled, we need to add up the distances traveled in each direction. So:Total distance traveled = 8 m + 4 m + 2 m + 3 m = 17 m. Therefore, the snake traveled a total of 17 meters.
2. To find the snake's final displacement, we need to find the total distance from the starting point to the ending point, in a straight line. We can use the Pythagorean theorem to do this:Final displacement = √((4 m - 2 m)² + (8 m - 3 m)²) = √(2² + 5²) = √29 ≈ 5.39 m. Therefore, the snake's final displacement was approximately 5.39 meters.
3. To find the snake's average speed, we can use the formula:Average speed = Total distance ÷ Time taken = 17 m ÷ 5 min = 3.4 m/min. We can convert this to meters per second (m/s) by dividing by 60 (since there are 60 seconds in a minute):Average speed = 3.4 m/min ÷ 60 = 0.057 m/s. Therefore, the snake's average speed was approximately 0.057 m/s.
4. To find the snake's average velocity, we need to find the total displacement and divide by the time taken. The direction of the displacement will be the same as the direction of the final displacement (i.e. from west to east, and from south to north). So:Total displacement = (4 m - 2 m, 8 m - 3 m) = (2 m, 5 m)Average velocity = Total displacement ÷ Time taken = (2 m, 5 m) ÷ 5 min = (0.4 m/min, 1 m/min)We can convert this to meters per second (m/s) by dividing by 60:Average velocity = (0.4 m/min ÷ 60, 1 m/min ÷ 60) = (0.0067 m/s, 0.017 m/s)Therefore, the snake's average velocity was approximately (0.0067 m/s, 0.017 m/s) in the east-north direction.
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An object moves along the x axis according to the equation x=2.70t
2
−2.00t+3.00, where x is in meters and t is in seconds. (a) Determine the average speed between t=1.80 s and t=3.00 s. m/s (b) Determine the instantaneous speed at t=1.80 s. m/s Determine the instantaneous speed at t=3.005. m/s (c) Determine the average acceleration between t=1.80 s and t=3.00 s. m/s
2
(d) Determine the instantaneous acceleration at t=1.805, m/s
2
Determine the instantaneous acceleration at t=3.00 s. m/s
2
(a) At what time is the object at rest?
(a) The average speed between t = 1.80 s and t = 3.00 s is approximately 9.186 m/s.
b) The instantaneous speed at t = 1.80 s is approximately 7.72 m/s.
the instantaneous speed at t = 3.005 s is approximately 14.53 m/s.
(c)The average acceleration between t = 1.80 s and t = 3.00 s is approximately 6.81 m/s².
d) The instantaneous acceleration at t = 1.805 s is approximately 5.40 m/s²
2) The object is at rest at approximately t = 0.370 s
a) We can find the displacement of the object by evaluating the position function at the given time points:
x₁ = 2.70(1.80)^2 - 2.00(1.80) + 3.00
x₂ = 2.70(3.00)^2 - 2.00(3.00) + 3.00
To find the total distance, we take the absolute difference between x₂ and x₁:
total_distance = |x₂ - x₁|
Finally, we divide the total distance by the time interval:
average_speed = total_distance / (3.00 - 1.80)
Let's calculate the average speed:
x₁ = 2.70(1.80)^2 - 2.00(1.80) + 3.00 ≈ 4.428
x₂ = 2.70(3.00)^2 - 2.00(3.00) + 3.00 ≈ 22.800
total_distance = |22.800 - 4.428| ≈ 18.372
average_speed = 18.372 / (3.00 - 1.80) ≈ 9.186 m/s
(b)the instantaneous speed at t = 1.80 s is approximately 7.72 m/s.
the instantaneous speed at t = 3.005 s is approximately 14.53 m/s.
v = dx/dt = d(2.70t^2 - 2.00t + 3.00)/dt
v = 5.40t - 2.00
v₁ = 5.40(1.80) - 2.00 ≈ 7.72 m/s
To determine the instantaneous speed at t = 3.005 s, we can follow the same process:
v₂ = 5.40(3.005) - 2.00 ≈ 14.53 m/s
(c)The average acceleration between t = 1.80 s and t = 3.00 s is approximately 6.81 m/s².
change_in_velocity = v₂ - v₁
average_acceleration = change_in_velocity / (3.00 - 1.80)
Let's calculate the average acceleration:
change_in_velocity = 14.53 - 7.72 ≈ 6.81 m/s
average_acceleration = 6.81 / (3.00 - 1.80) ≈ 6.81 m/s²
(d) the instantaneous acceleration at t = 1.805 s is approximately 5.40 m/s²
a = dv/dt = d(5.40t - 2.00)/dt
a = 5.40
To determine the instantaneous acceleration at t = 3.00 s, we can assume that the acceleration is constant since the velocity function is linear. In this case, the acceleration remains the same as the previous result:
a = 5.40 m/s²
Therefore, the instantaneous acceleration at t = 3.00 s is approximately 5.40 m/s².
2) The object is at rest at approximately t = 0.370 s.
v = 5.40t - 2.00 = 0
5.40t = 2.00
t = 2.00 / 5.40 ≈ 0.370 s
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Two aluminium components are bolted together using a copper pin. The diameter of the pin is 5.02 mm and the diameter of the hole is 5.00 mm. Given the thermal expansion coefficient of titanium 17× 10 −6∘C −1and aluminium 20×10 −6∘ C −1Determine the temperature when the components will separate.
The temperature when the components will separate is 219°C.
When two aluminum components are bolted together using a copper pin and the diameter of the pin is 5.02 mm while the diameter of the hole is 5.00 mm. The temperature when the components will separate can be determined as follows:
Given that;
The diameter of the pin is Dp = 5.02 mm
The diameter of the hole is Dh = 5.00 mm
The thermal expansion coefficient of titanium αt = 17 × 10⁻⁶/°C
The thermal expansion coefficient of aluminum αa = 20 × 10⁻⁶/°C
We know that the change in length is equal to the original length multiplied by the thermal expansion coefficient (α) and the change in temperature (ΔT).
Hence, the change in length of the copper pin ΔLp is given as;ΔLp = αt × Lp × ΔT …….(i)
Where Lp is the original length of the copper pin and ΔT is the change in temperature.Similarly, the change in length of the hole is ΔLh is given as;
ΔLh = αa × Lh × ΔT …….(ii)
Where Lh is the original length of the hole and ΔT is the change in temperature.
Since the copper pin and the hole are bolted together, the total change in length will be the difference in the change in length of the copper pin and the change in length of the hole. That is;
ΔL = ΔLp - ΔLh
ΔL = αt × Lp × ΔT - αa × Lh × ΔT
ΔL = ΔT (αt × Lp - αa × Lh) …….(iii)
Also, the total change in length is equal to the difference between the increase in the diameter of the copper pin and the decrease in the diameter of the hole. That is;
ΔL = Dp - Dh
ΔL = 5.02 - 5.00
ΔL = 0.02 mm
Substituting equation (iii) into the above equation;
ΔT (αt × Lp - αa × Lh) = 0.02
ΔT = 0.02 / (αt × Lp - αa × Lh)
ΔT = 0.02 / [(17 × 10⁻⁶ × Lp) - (20 × 10⁻⁶ × Lh)]
Substituting the values of Lp and Lh;
Lp = π/4 × Dp²
Lp = π/4 × (5.02)²
Lp = 19.81 mm
Lh = π/4 × Dh²
Lh = π/4 × (5.00)²
Lh = 19.63 mm
ΔT = 0.02 / [(17 × 10⁻⁶ × 19.81) - (20 × 10⁻⁶ × 19.63)]
ΔT = 2.19 × 10² °C
ΔT = 219 °C
Therefore, the temperature when the components will separate is 219°C.
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