We get the value of acceleration a = (-1.60×10^-19 C)(6.2×10^6 m/s)(1.6 T)sin(60°) / (9.11×10^-31 kg)
To calculate the magnitude of the acceleration of the electron, we can use the equation:
F = qvBsinθ
Where:
F = magnetic force on the electron
q = charge of the electron = -1.60×10^-19 C (negative because the electron has a negative charge)
v = velocity of the electron = 6.2×10^6 m/s
B = magnetic field strength = 1.6 T
θ = angle between the velocity vector and the magnetic field vector = 60°
The magnitude of the acceleration can be obtained using Newton's second law:
F = ma
Since F = qvBsinθ, we can rewrite the equation as:
ma = qvBsinθ
Solving for acceleration (a):
a = (qvBsinθ) / m
Substituting the given values:
a = (-1.60×10^-19 C)(6.2×10^6 m/s)(1.6 T)sin(60°) / (9.11×10^-31 kg)
Calculating this expression will give you the magnitude of the acceleration of the electron.
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A 100 watt light bulb is left on for two days (48 h). If electricity cost 12 per kWh, how much did it cost to run this light for two days?
To calculate the cost of running a 100-watt light bulb for two days, we need to determine the total energy consumption in kilowatt-hours (kWh) and then multiply it by the electricity cost per kWh.
Energy (kWh) = Power (kW) × Time (h)
Since the power of the light bulb is given in watts, we need to convert it to kilowatts:
Power (kW) = Power (W) / 1000
Power (kW) = 100 W / 1000 = 0.1 kW
Time (h) = 48 h
Energy (kWh) = 0.1 kW × 48 h = 4.8 kWh
Cost = Energy (kWh) × Cost per kWh
Given the cost per kWh is $0.12:
Cost = 4.8 kWh × $0.12/kWh = $0.576
Therefore, it cost $0.576 to run the 100-watt light bulb for two days (48 hours) at an electricity rate of $0.12 per kWh.
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A race car drives counter-clockwise around a circular track at a constant speed. When the car is at the easternmost point of the track, what is the direction of the car's velocity and acceleration vector? A) Velocity points north, Acceleration points north B) Velocity points north, Acceleration points west C) Velocity points west, Acceleration points west D) Velocity points west, Acceleration points south E) Velocity points north, Acceleration points south
When the car is at the easternmost point of the track, the direction of the car's velocity and acceleration vector is given by option E, Velocity points north, Acceleration points south.
A race car drives counter-clockwise around a circular track at a constant speed. When the car is at the easternmost point of the track, the direction of the car's velocity and acceleration vector is as follows: The direction of the velocity vector is tangent to the circular path. This means that when the car is at the easternmost point of the track, the direction of the velocity vector would be west, and it would be perpendicular to the radius of the circular path.
The direction of the acceleration vector is towards the center of the circular path. This is because any object that moves in a circular path, experiences centripetal acceleration directed towards the center of the path. Hence, when the car is at the easternmost point of the track, the direction of the acceleration vector would be towards the center of the circular path, which would be south. Therefore, the correct option would be E) Velocity points north, Acceleration points south.
Note: Please note that the length of the acceleration vector is given by:v = (v²)/r where v is the magnitude of the velocity vector and r is the radius of the circular path.
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How many collisions is required for a 2.0Mev neutron to be braked to an energy of 1/25eV on average, if the neutron moderator were (a) carbon, (b) iron? ANSWER (a) 112.1 Collisions (b) 501.5 Collisions 2. What is the moderating reason for nickel if it has ansorption cross-section of 4.8 and a dispersion cross-section of 17.5? How many collisions will it take to thermalize a 1.0Mev neutron? ANSWER 0.122:508 Collisions
1.a) Carbon:Given,Energy of the neutron, Ei = 2 MeV Target energy, Ef
= 1/25 eV
= 0.04 eV Neutron mass, m
= 1.67 × 10-27 kg Avogadro number, NA
= 6.023 × 1023 Moderating ratio, Mr
= m/(A × NA) where A is the atomic weight of the target Carbon atomic weight
= 12.01uMr for carbon
= 1.67 × 10-27/(12.01 × 1.66 × 10-27 × 6.023 × 1023)Mr
= 0.10003Let N be the number of collisions required to reduce the energy from Ei to Ef. The energy of the neutron is reduced by a factor of 2 in every collision.Energy after the first collision = 2/2
= 1.0 MeV Energy after the second collision
= 1/2
= 0.5 MeV Energy after the third collision
= 0.25 MeV Energy after the fourth collision
= 0.125 MeVEnergy after the Nth collision
= 2-N MeV The energy after the Nth collision must be equal to Ef.
Therefore,2-N = 0.04/1000eV2-N
= 1.6 × 10-8 Dividing both sides by 2N,2-N/N
= 1.6 × 10-8/N- log2
= -logN + log(1.6 × 10-8)- logN
= -log2 + log(1.6 × 10-8)logN
= log(1.6 × 10-8)/log2N
= 112.11.b) Iron:Mr for iron
= 1.67 × 10-27/(55.85 × 1.66 × 10-27 × 6.023 × 1023)Mr for iron
= 0.0333 The number of collisions required to reduce the energy from Ei to Ef will be the same as that for carbon because the difference between the moderating ratios is less than a factor of 2.Therefore, the number of collisions required = 112.1 collisions.2. For Nickel:The moderating reason for nickel is scattering.The total cross-section, σ = absorption cross-section, σa + scattering cross-section, σsσs = σ - σaσs
= 17.5 bσa
= 4.8 b Total mass of the nickel target, m
= A × NA × 1.67 × 10-27 where A is the atomic weight of nickel.The atomic weight of nickel
= 58.7u Therefore,m
= 58.7 × 1.66 × 10-27 × 6.023 × 1023
= 9.914 × 10-23 kg Mass of a single nickel atom
= 58.7 × 1.66 × 10-27 kg/6.023 × 1023
= 9.747 × 10-26 kg Let Ni be the number of collisions required to reduce the energy from 1.0 MeV to 0.04 eV.
Then, the energy of the neutron after the Nth collision, EN = 1.0/2N Me V.The neutron is assumed to be scattered at every collision. The scattered neutron is assumed to be isotropic, i.e., it is equally likely to be scattered in any direction. Let dΩ be the solid angle subtended by the neutron detector. The probability of the scattered neutron being detected is proportional to dΩ. Therefore, let the probability of detection be dΩ/4π.The mass number of nickel is 59. Therefore, the moderating ratio is Mr = 9.747 × 10-26/(59 × 1.66 × 10-27 × 6.023 × 1023)Mr = 0.0275 The scattering cross-section is related to the moderating ratio and the total cross-section as follows:σs = 4πMr2NA/m The value of σs is calculated to be 8.7 b.The scattering cross-section is related to the number of collisions required to reduce the energy from 1.0 MeV to 0.04 eV as follows:Ni = ln(1.0/0.04)/σsNi = 508.11 or 0.122 collisions (rounded off to three decimal places).
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Which of the following are not true about the surface tension?
A. Small Liquid Drop take spherical shape due to surface tension
B. Surface tension is a vector quantity
C. The surface tension of a liquid is a molecular phenomenon
D. The surface tension of liquid depends on length but not on the area
The surface tension of liquid depends on length but not on the area is not true about the surface tension.
The given statement is incorrect.
Surface tension is defined as the energy needed to increase the surface area of a liquid. In other words, the amount of force required to break the surface film and expose more surface area to the surrounding environment is the surface tension. It is measured per unit length. The dimensions of surface tension are energy per unit area, which can be expressed as Newtons/meter or dynes/cm.In a simple way, surface tension can be explained as the force that exists in the surface layer of a liquid because of the intermolecular forces of attraction. Surface tension is a molecular phenomenon. Surface tension is a scalar quantity, which means that it has no direction. Small liquid drops take a spherical shape due to surface tension. This is because the surface tension of liquids has the effect of minimizing the surface area of the liquid, which causes it to form a sphere. Therefore, option D is incorrect.
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A basketball leaves a player's hands at a height of 2.10 m above the floor. The basket is 3.05 m above the floor. The player likes to shoot the ball at a 38.0 degree angle.
1. If the shot is made from a horizontal distance of 7.40 m and must be accurate to ±0.22m±0.22m (horizontally), what is the range of initial speeds allowed to make the basket?
Express your answers using thre significant figures separated by a comma.
To calculate the range of initial speeds allowed to make the basket, we can use the projectile motion equations. The range of initial speeds allowed to make the basket is approximately 1.870 m/s.
To calculate the range of initial speeds allowed to make the basket, we can use the projectile motion equations.
The horizontal range (R) of a projectile can be calculated using the equation:
R = (v^2 * sin(2θ)) / g
Where:
R is the range,
v is the initial speed of the projectile,
θ is the launch angle, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the launch angle (θ) is given as 38.0 degrees, the horizontal distance (±0.22 m) is given as the range, and we need to find the range of initial speeds (v).
Rearranging the equation, we can solve for v:
v = sqrt((R * g) / sin(2θ))
Now we substitute the given values:
R = ±0.22 m
g = 9.8 m/s^2
θ = 38.0 degrees
Calculating the range of initial speeds:
v = sqrt((±0.22 m * 9.8 m/s^2) / sin(2 * 38.0 degrees))
v = sqrt(2.156 m^2/s^2 / 0.615661 m^2/s^2)
v = sqrt(3.5017)
v ≈ 1.870 m/s
Therefore, the range of initial speeds allowed to make the basket is approximately 1.870 m/s.
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Four objects each with charge +1.8×10−7C are located at the comers of a square whose sides are 2.3 m long. Part C Find the total electric potential energy of the system consisting of the four charged objects. Express your answer with the appropriate units.
The total electric potential energy of the system consisting of the four charged objects is 7.26×10⁻¹⁰ J.
The electric potential energy of a collection of point charges is defined by
U = (1/2) q1q2 / (4πεo r12)
The total electric potential energy is the sum of all of the electric potential energy of all the pairs of charges in the system.
To find the total electric potential energy, we need to calculate the electric potential energy of each pair of charges, and then add up the result.Each charge interacts with every other charge in the system, so there are 6 distinct pairs of charges.
Each charge has the same magnitude, so we can write q1 = q2 = q3 = q4 = q.
To find the distance r between each pair of charges, we can use the Pythagorean theorem:
r²= (2.3/2)² + (2.3/2)² = 1.62 m
Now, we can calculate the electric potential energy of each pair of charges:
U₁₂ = U₁₃ = U₁₄= U₂₃ = U₂₄= U₃₄ = (1/2) (1.8×10−7)² / (4πεo (1.62))
U₁₂ = U₁₃ = U₁₄= U₂₃ = U₂₄= U₃₄ = 1.21×10⁻¹⁰ J
Finally, we can add up all of the electric potential energy of all the pairs of charges in the system to find the total electric potential energy.
Utotal = U₁₂ + U₁₃ + U₁₄ + U₂₃ + U₂₄ + U₃₄
Utotal = 6 (1.21×10⁻¹⁰) J
Utotal = 7.26×10⁻¹⁰ J.
Therefore, the total electric potential energy of the system consisting of the four charged objects is 7.26×10⁻¹⁰ J.
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7-2 A freight train has a mass of The wheels of the locomotive push back on the tracks with a constant net force of 7.50×10
5
N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h ? s(±2s)
It would take approximately 44.4 seconds to increase the speed of the train from rest to 80.0 km/h.
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
Given:
Mass of the freight train (m) = 1.50 × 10⁶ kg
Net force acting on the locomotive (F) = 7.50 × 10⁵ N
Final speed of the train (v) = 80.0 km/h = 80.0 km/h * (1000 m/km) / (3600 s/h) ≈ 22.2 m/s
We can assume that the acceleration of the train is constant during this time interval.
Using Newton's second law, we have:
F = m * a
Solving for acceleration (a):
a = F / m
Substituting the known values:
a = (7.50 × 10⁵ N) / (1.50 × 10⁶ kg) = 0.5 m/s^2
Now, we can use the kinematic equation to find the time it takes for the train to reach its final speed from rest:
v = u + at
Where:
u is the initial velocity (0 m/s),
v is the final velocity (22.2 m/s),
a is the acceleration (0.5 m/s²),
t is the time.
Substituting the known values:
22.2 m/s = 0 + (0.5 m/s²) * t
Solving for time (t):
t = (22.2 m/s) / (0.5 m/s²) ≈ 44.4 seconds.
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A motorist sees a reduced speed sign and quickly reduces her speed to 10 m/s. If the car decelerates at a rate of −5.0 m/s
2
in 0.1 min, what was her initial speed? 8. As a driver pulls onto the freeway, she increases her speed from 5.0 m/s to 35.0 m/s in 4.5 s. What is her acceleration? 9. A swan on a lake gets airborne by flapping the swan's wings and running on top of the water. If the swan must reach a velocity of 2.0 m/s to take off and accelerates from rest at an average rate of 0.50 m/s
2
, how far will the swan travel before becoming airborne? 10. Some rabbits can jump as high as four feet. Calculate the rabbit's vertical speed when the rabbit leaves the ground. (Hint: Use 32.2ft/s
2
instead of 9.81 m/s
2
. Also, what is the rabbit's speed at the top of the jump?) 11. There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. What velocity will it be going when it strikes the ground? (Hint: Assume upward is the positive direction.)
1. The initial speed of the car was 40 m/s.
2. The acceleration of the driver was approximately 6.67 m/s^2.
3. The swan will travel a distance of 4.0 meters before becoming airborne.
4. The velocity of the boulder when it strikes the ground is approximately 70 m/s.
1. To find the initial speed of the car, we can use the formula for constant acceleration:
v = u + at
Where:
v = final velocity (10 m/s, as given)
u = initial velocity (unknown)
a = acceleration (-5.0 m/s^2, as given)
t = time (0.1 min, which needs to be converted to seconds)
First, let's convert the time from minutes to seconds:
0.1 min = 0.1 * 60 s = 6 s
Now we can plug the values into the formula and solve for u:
10 m/s = u + (-5.0 m/s^2) * 6 s
10 m/s = u - 30 m/s
Rearranging the equation to solve for u:
u = 10 m/s + 30 m/s
u = 40 m/s
Therefore, the initial speed of the car was 40 m/s.
2. To find the acceleration of the driver, we can use the formula for acceleration:
a = (v - u) / t
Where:
a = acceleration (unknown)
v = final velocity (35.0 m/s, as given)
u = initial velocity (5.0 m/s, as given)
t = time (4.5 s, as given)
Plugging in the values:
a = (35.0 m/s - 5.0 m/s) / 4.5 s
a = 30.0 m/s / 4.5 s
a ≈ 6.67 m/s^2
Therefore, the acceleration of the driver was approximately 6.67 m/s^2.
3. To find the distance traveled by the swan before becoming airborne, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (2.0 m/s, as given)
u = initial velocity (0 m/s, as the swan starts from rest)
a = acceleration (0.50 m/s^2, as given)
s = distance (unknown)
Plugging in the values:
(2.0 m/s)^2 = (0 m/s)^2 + 2 * 0.50 m/s^2 * s
4.0 m^2/s^2 = 1.0 m^2/s^2 * s
s = 4.0 m^2/s^2 / 1.0 m^2/s^2
s = 4.0 m
Therefore, the swan will travel a distance of 4.0 meters before becoming airborne.
4. To find the velocity of the boulder when it strikes the ground, we can use the equation for free fall:
v^2 = u^2 + 2gh
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s, as the boulder starts from rest)
g = acceleration due to gravity (9.81 m/s^2)
h = height (250 m, as given)
Plugging in the values:
v^2 = (0 m/s)^2 + 2 * 9.81 m/s^2 * 250 m
v^2 = 0 m^2/s^2 + 4905 m^2/s^2
v^2 = 4905 m^2/s^2
v ≈ √4905 m/s
v ≈ 70 m/s
Therefore, the velocity of the boulder when it strikes the ground is approximately 70 m/s.
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An astronaut on a strange new planet finds that she can jump up to a maximum height of 28 meters when her initial upward speed is 5.1 m/s. What is the magnitude of the acceleration of gravity on the planet? Answer in units of m/s2 and round to two decimal places.
An astronaut on a strange new planet finds that she can jump up to a maximum height of 28 meters when her initial upward speed is 5.1 m/s. The magnitude of the acceleration of gravity on the planet is 0.46 m/s².
To find the magnitude of the acceleration of gravity on the planet, we can use the kinematic equation for vertical motion:
Δy = v₀y² / (2 * g)
where Δy is the change in height, v₀y is the initial upward velocity, and g is the acceleration due to gravity.
Given:
Δy = 28 meters
v₀y = 5.1 m/s
We can rearrange the equation to solve for g:
g = v₀y² / (2 * Δy)
Plugging in the values, we have:
g = (5.1 m/s)² / (2 * 28 meters)
g = 26.01 m²/s² / 56 meters
g ≈ 0.464 m/s²
Therefore, the magnitude of the acceleration of gravity on the planet is approximately 0.46 m/s².
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A defibrillator containing a 0.2 F capacitor is used to shock the heart of a patient by holding it to the patient chest. if the defibrillator stores an energy of 2000 J, what is the charge Q on each plate of the defibrillator ?
If the defibrillator stores an energy of 2000 J,the charge on each plate of the defibrillator is 40 C.
The energy stored in a capacitor can be calculated using the equation:
E = (1/2) * C * [tex]V^2[/tex]
Where:
E is the energy stored in the capacitor (2000 J in this case).
C is the capacitance of the capacitor (0.2 F).
V is the voltage across the capacitor.
We can rearrange the equation to solve for the voltage:
V = sqrt((2 * E) / C)
Substituting the given values:
V = sqrt((2 * 2000) / 0.2)
V = sqrt(40000)
V = 200 V
The charge Q on each plate of the capacitor can be calculated using the equation:
Q = C * V
Substituting the values:
Q = 0.2 * 200
Q = 40 C
Therefore, the charge on each plate of the defibrillator is 40 C.
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The energy stored in a defibrillator gives us the necessary variables to calculate the charge on each 'plate' of the defibrillator. By using physics formulas relating to capacitors, we find that the charge on each plate is 2000 Coulombs.
Explanation:The subject of your question involves the concepts of Physics, specifically relating to the use of capacitors in defibrillators. We know that a capacitor stores energy, particularly in a defibrillator which sets pulses of electricity into the heart to keep it beating regularly.
The energy stored in a capacitor is given by the formula U = 0.5 * C * V^2, where U represents energy (in joules), C is the capacitance (in farads) and V is the voltage across the capacitor (in volts). From your question, we're given that U is 2000 J and C is 0.2 F. However, we need to calculate V, the voltage across the capacitor, which is necessary in order to find the charge Q on each plate of the defibrillator.
To find V, we'll rearrange our energy equation solving for V: V = sqrt((2 * U) / C). Substituting our given values, we get V = sqrt((2 * 2000 J) / 0.2 F) = 10000 V.
Lastly, we can find the charge Q using the formula Q = C * V. Substituting our previous values, we get Q = (0.2 F) * (10000 V) = 2000 Coulombs. So the charge on each plate of the defibrillator is 2000 C.
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A driver's reaction time is approximately 0.4 s. How many m does a car travel in a school zone before the average driver can react? 9. A fast kid can run 1/4 the speed of the fastest high school runner. How far can a fast kid run during a driver's reaction time? 10. The circumference of the earth is 40,000 km. If you are resting on a beach at the equator, how fast are you moving in m/s around the earth's axis of rotation? 11. On a beach at the equator, how far do you move during a driver's reaction time relative to the beach? 12. Relative to the center of the earth?
During the driver's reaction time, you would move approximately 463.0 meters relative to the center of the Earth.
To find the distance a car travels in a school zone before the average driver can react, we need to calculate the distance covered during the reaction time.
Since the reaction time is approximately 0.4 seconds, we can use the formula:
Distance = Speed × Time
Assuming the car is stationary, the distance covered during the reaction time is:
Distance = 0.4 s × 0 m/s = 0 m
Therefore, the car doesn't travel any distance during the average driver's reaction time.
If a fast kid can run 1/4 the speed of the fastest high school runner, we need to know the speed of the fastest high school runner to calculate the distance the fast kid can run during the reaction time.
Without that information, we cannot provide a specific answer.
On a beach at the equator, the distance you move during a driver's reaction time relative to the beach depends on your initial speed. If you are standing still, the distance covered will be zero.
Relative to the center of the Earth, the distance you move during a driver's reaction time is determined by the rotational speed of the Earth. Since the reaction time is relatively short, the distance covered will be very small.
To calculate the distance, we can use the formula:
Distance = Speed × Time
The speed can be calculated by dividing the circumference of the Earth by the length of a day (24 hours or 86,400 seconds).
Using a circumference of 40,000 km (40,000,000 m), we get:
Speed = 40,000,000 m / 86,400 s = 463.0 m/s (approximately)
Thus, during the driver's reaction time, you would move approximately 463.0 meters relative to the center of the Earth.
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A certain aircraft has a liftoff speed of 130 km/h. (a) What minimum constant accelefation does the aircraft require if it is to be airborne after a takeoff run of 245 m ? m/s
2
(b) How long does it take the aircraft to become airborne?
The time taken by the aircraft to become airborne is 13.71 s.
Initial Velocity (u) = 0 m/s,
Final Velocity (v) = 130 km/h = (130 × 5/18) m/s = 36.11 m/s,
Distance covered (s) = 245 m(a)
We need to find the minimum constant acceleration required by the aircraft to be airborne after a takeoff run of 245 m.
Using 2nd equation of motion:
v² = u² + 2as36.11² = 0 + 2a(245)a = (36.11)²/ (2 × 245)a = 2.63 m/s²
Therefore, the minimum constant acceleration required by the aircraft to be airborne after a takeoff run of 245 m is 2.63 m/s².
(b) We need to find the time taken by the aircraft to become airborne.
Using 3rd equation of motion:v = u + at36.11 = 0 + a × tt = 36.11/a= 36.11/2.63t = 13.71 s
Therefore, the time taken by the aircraft to become airborne is 13.71 s.
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The voltage across a capacitor with capacitance C=10μF is v
c
(t)=4.8−4.8e
−100t
V for t≥0s. Sketch the capacitor voltage and current on separate axes. The sketch should clearly indicate the time scale and the voltage and current values.
The voltage across a capacitor with capacitance C=10μF is given by the equation v_c(t) = 4.8 - 4.8e^(-100t) V for t≥0s.To sketch the capacitor voltage and current on separate axes, let's first focus on the voltage.
The given equation for the voltage across the capacitor is v_c(t) = 4.8 - 4.8e^(-100t) V.
We can see that the initial voltage across the capacitor is 4.8V (when t=0s), and as time increases, the voltage decreases exponentially. The exponential term e^(-100t) is responsible for the decay of the voltage.
To sketch the voltage on a graph, we need to determine the time scale and the voltage values.
Let's consider a time range of t=0s to t=1s. We can divide this time range into smaller intervals, say t=0s, t=0.2s, t=0.4s, t=0.6s, t=0.8s, and t=1s.
At t=0s, the voltage is v_c(0) = 4.8V. At t=0.2s, the voltage is v_c(0.2) = 4.8 - 4.8e^(-100*0.2) V. Similarly, we can calculate the voltage for the other time intervals.
Plotting these voltage values on the y-axis against the corresponding time intervals on the x-axis, we can sketch the capacitor voltage.
Now let's consider the current through the capacitor. The current is given by the equation i_c(t) = C(dv_c(t)/dt), where C is the capacitance (10μF).
To sketch the current on a separate axis, we need to calculate the derivative of the voltage with respect to time, and then multiply it by the capacitance C.
Taking the derivative of v_c(t) = 4.8 - 4.8e^(-100t) V with respect to t, we get dv_c(t)/dt = 480e^(-100t) A/s. Multiplying this by the capacitance C=10μF, we get the current i_c(t) = 4800e^(-100t) A/s.
Using the same time intervals as before, we can calculate the current values for each interval and plot them on a separate graph.
Remember to clearly indicate the time scale on the x-axis and the voltage/current values on the y-axis for both graphs.
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A keen physics student starts running at 3.4 m/s and accelerates at 0.55 for 3.7 s. What is the final velocity of the keen physics student?
Sure, I can help you with that. To find the final velocity of the keen physics student, you need to use the formula for final velocity under constant acceleration:
[tex]$$v = u + at$$[/tex]
where:
- [tex]$v$[/tex] is the final velocity
- [tex]$u$[/tex] is the initial velocity
- [tex]$a$[/tex] is the acceleration
- [tex]$t$[/tex] is the time
Plugging in the given values, we get:
[tex]$$v = 3.4 + 0.55 \times 3.7$$[/tex]
[tex]$$v = 5.435$$[/tex]
Therefore, the final velocity of the keen physics student is 5.435 m/s I hope this helps!
(35pts.) A block of mass \( m_{1}=m \) sits (without sliding) on top of a bigger block of mass \( m_{2}=2 m \), which is connected through an ideal pulley to a third block of mass \( m_{3}=3 m \) that
The force acting on the third block can be found using the equations of motion. Let T be the tension in the string. The block of mass m1 experiences only one force acting on it, i.e., its weight (mg), which acts in the downward direction.
Therefore, the normal force exerted on it by the bigger block must be equal and opposite to its weight. Hence, the force acting on the bigger block in the upward direction is (2m)(g) + T, where g is the acceleration due to gravity.The force acting on the third block in the downward direction is 3mg, where g is the acceleration due to gravity. Therefore, using Newton's second law of motion, we can write:
3mg = T - (2m)(g)
Therefore, T = 5mg Now, using T = 5mg, we can find the acceleration of the system as follows:
5mg - (2m)(g) = (5m + 2m)a3g - 2g = 7ma = g/7
Therefore, the acceleration of the system is g/7. In this problem, we have three blocks of masses m1, m2, and m3 connected by a string passing over an ideal pulley. The first block is placed on top of the second block, which is then connected to the third block through the string passing over the pulley. The question asks us to find the acceleration of the system and the tension in the string.Let's first consider the forces acting on each block. The block of mass m1 experiences only one force acting on it, i.e., its weight (mg), which acts in the downward direction. Therefore, the normal force exerted on it by the bigger block must be equal and opposite to its weight. Hence, the force acting on the bigger block in the upward direction is (2m)(g) + T, where g is the acceleration due to gravity. Similarly, the force acting on the third block in the downward direction is 3mg, where g is the acceleration due to gravity.Now, using Newton's second law of motion, we can write:
3mg = T - (2m)(g)
Therefore, T = 5mg Now, using T = 5mg, we can find the acceleration of the system as follows:
5mg - (2m)(g) = (5m + 2m)a3g - 2g = 7ma = g/7
Therefore, the acceleration of the system is g/7.
In conclusion, we have found the acceleration of the system to be g/7 and the tension in the string to be 5mg.
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Electrons move through a certain electrical circuit at an average speed of 2.4×10
−2
m/s. How long (in minutes) does it take an electron to traverse a 2.0-m light bulb filament coil?
It takes approximately 1.39 minutes for an electron to traverse a 2.0-m light bulb filament coil at an average speed of 2.4 x 10^-2 m/s.
The speed of electrons in a circuit is given by the current. This speed is expressed in meters per second (m/s). We can calculate the time it takes an electron to traverse a given distance in a circuit using the formula: time = distance/speed. In this case, we are given the speed of electrons, and we need to find the time it takes them to traverse a 2.0-m light bulb filament coil.
We can use the formula: time = distance/speed
Distance = 2.0 m
Speed = 2.4 x 10^-2 m/s
Substituting these values in the formula above, we get:
time = distance/speed
time = 2.0 m / (2.4 x 10^-2 m/s)
time = 83.3 s
Therefore, it takes 83.3 seconds for an electron to traverse a 2.0-m light bulb filament coil. To convert seconds to minutes, we can divide the answer by 60:
83.3 s / 60 s/min = 1.39 min (rounded to two decimal places)
Therefore, it takes approximately 1.39 minutes for an electron to traverse a 2.0-m light bulb filament coil at an average speed of 2.4 x 10^-2 m/s.
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Suppose a 59.5 kg gymnast climbs a rope. (a) What is the tension (in N) in the rope if he climbs at a constant speed? N (b) What is the tension (in N ) in the rope if he accelerates upward at a rate of 1.55 m/s
2
?
a) When the gymnast climbs at a constant speed, the tension in the rope is equal to the weight of the gymnast.
Therefore,
T = mg
= 59.5 kg × 9.81 m/s²
= 583.695 N.
The tension in the rope when the gymnast climbs at a constant speed is 583.695 N.
b) When the gymnast accelerates upward at a rate of 1.55 m/s², the tension in the rope will be greater than his weight.
T = mg + ma
= 59.5 kg × 9.81 m/s² + 59.5 kg × 1.55 m/s²T
= 584.475 N + 92.725 N
= 677.2 N.
The tension in the rope when the gymnast accelerates upward at a rate of 1.55 m/s² is 677.2 N.
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8. A ball is thrown upward and returns to the person's hand at the same level as it was thrown. If this took 4.0 s : a. How long did it take to get to the highest point? b. How fast was it thrown up? c. How high did it go?
A ball thrown up vertically reaches a maximum height of 19.2 m and takes 2.0 seconds to get to the highest point.
a. How long did it take to get to the highest point?
The time it takes to get to the highest point is half of the total time, so it took 2.0 seconds to get to the highest point.
b. How fast was it thrown up?
We can use the following equation to find the initial velocity of the ball:
v = u + at
where:
v is the final velocity (0 m/s, since the ball comes to rest at the highest point)
u is the initial velocity
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time it takes to reach the highest point (2.0 seconds)
0 = u - 9.8 * 2.0
u = 19.6 m/s
c. How high did it go?
The maximum height the ball reached is given by the following equation:
h = u^2 / 2g
where:
h is the maximum height
u is the initial velocity (19.6 m/s)
g is the acceleration due to gravity (-9.8 m/s^2)
h = (19.6)^2 / 2 * -9.8
h = 19.2 m
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A 1500 kg elevator, suspended by a single cable with tension 16.0kN, is measured to Part C be moving upward at 2.6 m/s. Air resistance is negligible. How much work does gravity do on the elevator as it rises 10.0 m ? Express your answer with the appropriate units. What is the elevator's kinetic energy after rising 10.0 m ? Express your answer with the appropriate units. What is the elevator's speed after rising 10.0 m ? Express your answer with the appropriate units.
Given data:
Mass of elevator (m) = 1500 kg
Tension (T) = 16.0 kN = 16000 N
Velocity (v) = 2.6 m/s
Displacement (s) = 10.0 m
Work done by gravity on the elevator can be calculated using the formula:
W = mgh
Where:
m = mass of the object
g = acceleration due to gravity
h = height of the object above the ground
Substituting the values of m, g, and h, we get:
W = mgh
= 1500 kg × 9.8 m/s² × 10.0 m
= 147000 J
Therefore, the work done by gravity on the elevator as it rises 10.0 m is 147000 J.
The kinetic energy of the elevator can be calculated using the formula:
KE = (1/2)mv²
Where:
m = mass of the object
v = velocity of the object
Substituting the values of m and v, we get:
KE = (1/2)mv²
= (1/2) × 1500 kg × (2.6 m/s)²
= 10140 J
Therefore, the kinetic energy of the elevator after rising 10.0 m is 10140 J.
The final velocity of the elevator can be calculated using the formula:
v = √(v₀² + 2as)
Where:
v₀ = initial velocity of the elevator
a = acceleration of the elevator = g = 9.8 m/s²
s = displacement of the elevator
Substituting the values of v₀, a, and s, we get:
v = √(v₀² + 2as)
= √(2.6 m/s)² + 2(9.8 m/s²)(10.0 m)
= √264
= 16.2 m/s
The elevator's speed after rising 10.0 m is 16.2 m/s.
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Children playing in a playground on the flat roof of a city school lose their ball to the parking lot below. One of the teachers kicks the ball back up to the children as shown in the figure below. The playground is 6.20 m above the parking lot, and the school building's vertical wall is h=7.50 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle θ=53.0
∘
above the horizontal at d.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (Due to the nature of this problem, do not intermediate values in your calculations-including answers submitted in WebAssign.) (a) Find the speed (in m/s ) at which the ball was launched. m/s (b) Find the vertical distance (in m) by which the ball clears the wall. & m (c) Find the horizontal distance (in m ) from the wall to the point on the roof where the ball lands. 28 m (d) What If? If the teacher always launches the ball with the speed found in part (a), what is the minimum angle (in degrees above the horizontal) at which he can launch the ball and still clear the playground railing? (Hint: You may need to use the trigonometric identity sec
2
(θ)=1+tan
2
(θ).) * ∘ above the horizontal (e) What would be the horizontal distance (in m ) from the wall to the point on the roof where the ball lands in this case?
(a) 14.9 m/s, (b) 3.8 m, (c) 28 m, (d) 25.3° above the horizontal, (e) 9.92 m. (a) Speed of the ball launched The initial velocity of the ball can be found out from the vertical motion equations. As the ball is launched at an angle θ above the horizontal, so the initial velocity of the ball will have two components: horizontal and vertical.Let u be the initial velocity, then, Vertical component: u * sin(53) = gt, where g is the acceleration due to gravity and t is the time taken by the ball to reach a point vertically above the wall. Hence, u = gt/sin(53) = (9.8 * 2.2)/sin(53) = 14.9 m/s. Horizontal component: u * cos(53) = d = distance of the point from the wall.
Hence, u = d/cos(53). Equating both the values of u, d/cos(53) = 14.9. Hence, d = 14.9 * cos(53) = 10.9 m.(b) Vertical distance cleared by the ballThe final velocity of the ball can be calculated using the vertical motion equations. v^2 = u^2 + 2gh, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the height from which the ball was launched. As the ball reaches the same height, so, 0 = u^2 + 2gh, v^2 = 2gh, v = sqrt(2gh), v = sqrt(2 * 9.8 * 6.2) = 11.1 m/s. Therefore, The vertical distance by which the ball clears the wall is, h = (v^2 - u^2)/2g = (11.1^2 - 14.9^2)/(2 * 9.8) = 3.8 m.(c) Horizontal distance from the wall to the point on the roof where the ball landsThe horizontal distance can be calculated using the horizontal motion equations. d = ut + (1/2)at^2, where a = 0 as there is no acceleration in the horizontal direction.
Hence, d = u * cos(53) * t = (14.9 * cos(53)) * 2.2 = 28 m.(d) Minimum angle The minimum angle can be found out by equating the height cleared by the ball to the height of the railing. Maximum height attained by the ball, h = (v^2 sin^2(θ))/(2g), h = (14.9^2 * sin^2(θ))/(2 * 9.8) Height of the railing, 1.3 m. Hence, (14.9^2 * sin^2(θ))/(2 * 9.8) = 1.3. Solving this, sin^2(θ) = 0.186, sin(θ) = 0.431, θ = 25.3°. Hence, the minimum angle is 25.3° above the horizontal.(e) Horizontal distance The horizontal distance can be calculated using the formula, d = (v^2 sin(2θ))/g. d = (14.9^2 sin(2 * 25.3°))/9.8 = 9.92 m. Hence, the horizontal distance from the wall to the point on the roof where the ball lands would be 9.92 m.
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A point charge q0 that has a charge of 0.800μC is at the origin. A second particle qthat has a charge of 2.00μC and a mass of 0.0800 g is placed at x=0.800 m. 1) What is the potential energy of this system of charges? (Express your answer to three significant figures.) mJ Your submissions: Computed value: 0.0180 Submitted: Wednesday, September 21 at 11:12AMFeedback: It appears you have made a power of ten error. 2) If the particle with charge q is released from rest, what will its speed be when it reaches x=2.00 m? (Express your answer to three significant figures.)
The potential energy of the system of charges is 17.98 mJ, and the speed of the particle when it reaches x=2.00 m is approximately 17.91 m/s.
1) Potential energy of the system:The potential energy (PE) between two point charges can be calculated using the formula:
PE = (k * |q₁ * q₂|) / r
where k is the electrostatic constant (k = 8.99 x 10⁹ N*m²/C²), q₁ and q₂ are the charges, and r is the separation between the charges.
In this case, the potential energy between the charges q₀ and q can be calculated as follows:
PE = (k * |q₀ * q|) / r
Substituting the given values:
PE = (8.99 x 1010⁹ N*m²/C²) * |(0.800 x 10⁻⁶ C) * (2.00 x 10⁻⁶ C)| / (0.800 m)
Calculating the value:
PE = 17.98 mJ
Therefore, the potential energy of the system is 17.98 millijoules (mJ).
2) Speed of the particle at x=2.00 m:To find the speed of the particle when it reaches x=2.00 m, we can use the conservation of energy. The initial potential energy (PE_initial) is equal to the final kinetic energy (KE_final).
PE_initial = KE_final
(8.99 x 10⁹ N*m²/C²) * |(0.800 x 10⁻⁶ C) * (2.00 x 10⁻⁶ C)| / (0.800 m) = (1/2) * m * v²
We need to find the speed v when x=2.00 m. Rearranging the equation:
v = √[(2 * PE_initial) / m]
Substituting the values:
v = √[(2 * 17.98 x 10⁻³ J) / (0.0800 x 10⁻³ kg)]
Calculating the value:
v ≈ 17.91 m/s
Therefore, the speed of the particle when it reaches x=2.00 m is approximately 17.91 m/s.
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A crate slides across a frictionless horizontal surface with constant acceleration 3.90 m/s
2
and over a time interval reaches a final velocity of 11.4 m/s. (a) If its initial velocity is 5.70 m/s, what is its displacement (in m ) during the time interval? (Indicate the direction with the sign of your answer.) m (b) What is the distance it travels (in m) during this interval? m (c) If its initial velocity is −5.70 m/s, what is its displacement (in m) during the time interval? (Indicate the direction with the sign of your answer.) m (d) What is the total distance it travels (in m) during the interval in part (c)? m
The displacement of the crate during the time interval is 18.3 m in the direction of acceleration. The distance traveled by the crate during this interval is 21.9 m. The total distance traveled by the crate during the interval in part (c) is 21.9 m.
(a) The displacement of the crate during the time interval is 18.3 m in the direction of acceleration.
The displacement of an object can be determined using the formula: displacement = (final velocity² - initial velocity²) / (2 * acceleration). Substituting the given values, we have (11.4 m/s)² - (5.70 m/s)² / (2 * 3.90 m/s²) = 18.3 m.
(b) The distance traveled by the crate during this interval is 21.9 m.
Distance traveled is the total length covered by an object, regardless of direction. In this case, we consider only the magnitudes of the velocities. Using the formula: distance = (final velocity - initial velocity) * time, we have (11.4 m/s - 5.70 m/s) * t = 21.9 m.
(c) If the initial velocity is -5.70 m/s, the displacement of the crate during the time interval is -18.3 m in the opposite direction of acceleration.
The negative initial velocity indicates that the crate is moving in the opposite direction. Using the same displacement formula as in (a), we obtain (-11.4 m/s)² - (-5.70 m/s)² / (2 * 3.90 m/s²) = -18.3 m.
(d) The total distance traveled by the crate during the interval in part (c) is 21.9 m.
Total distance considers the magnitude of the distance traveled without considering the direction. Therefore, it is the same as the answer in part (b) since distance does not depend on the initial velocity.
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I have attempted several times and have not been right. The answer is not 0.00548. It could be a rounding error.
The table I have been given suggests using 19 x 10^-6 as the coefficient of linear expansion of brass (alpha). Thank you in advance!!A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum is 3.68 s. The temperature rises by 149C
∘
, and the length of the wire increases. Determine the change in the period of the heated pendulum. Number Units
The change in the time period of the heated pendulum is 0.00096 seconds (rounded off to 3 significant figures).
The change in the period of a heated pendulum can be determined using the following equations:
L = L₀ (1 + αΔT) (1)
T = 2π√(L/g) (2)
Where L₀ is the length of the wire at the initial temperature, α is the coefficient of linear expansion of brass, ΔT is the temperature change, T is the time period of the pendulum, and g is the acceleration due to gravity.
By substituting equation (1) into equation (2), we can express the time period in terms of L₀ and ΔT:
T = 2π√((L₀ (1 + αΔT))/g) = 2π(L₀/g)√(1 + αΔT)
Hence, the change in the period is given by:
ΔT = T - T₀ = 2π(L₀/g)√(1 + αΔT) - 2π(L₀/g)
Squaring both sides, we get:
ΔT² = (2πL₀/g)²(1 + αΔT) - (2πL₀/g)²
Simplifying further, we obtain:
ΔT = [(2πL₀/g)²αΔT] / [1 - (2πL₀/g)²]
Substituting the given values, we find:
ΔT = [(2π x 19.00 x 10^-6 x 9.81 x 3.68²) / (1 - (2π x 19.00 x 10^-6 x 9.81 x 3.68²))]
After calculating, we round off the result to 3 significant figures:
ΔT = 0.00096 s
Therefore, the change in the time period of the heated pendulum is 0.00096 s.
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A car is traveling at speed v
0
on a straight road. A traffic light at distance d turns yellow. It takes a second for the driver to apply the brakes, and then the car has constant (negative) acceleration a. Determine (A) the acceleration required to stop at the light and (B) the stopping time. [ Data : v
0
=56 km/hr;d=44 m ]
Given, Initial velocity of the car = v₀ = 56 km/hr = 15.56 m/sDistance to travel = d = 44 m Time taken to stop = t Acceleration of the car is given by the formula:
v = u + at Here, the final velocity is 0, the initial velocity is v₀, acceleration is a, and time taken is t. Rearranging the above equation, we get, t = (v - u) / a When the car comes to rest, the distance traveled is given by:s = (u² - v²) / 2aHere, the final velocity is 0, the initial velocity is v₀, distance is d and acceleration is a. Substituting the values, we get; d = (v₀² - 0²) / (2a)44 = (15.56²) / (2a)
Solving for 'a', we get a = 60.7 m/s²Therefore, the acceleration required to stop at the traffic light is 60.7 m/s².Now, using the formula obtained above,t = (v - u) / a= (0 - 15.56) / (-60.7) = 0.255 seconds. Therefore, the stopping time of the car is 0.255 seconds.
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Two vectors,
r
and
s
lie in the xy plane. Their magnitudes are 3.78 and 4.81 units, respectively, and their directions are 317
∘
and 50.0
∘
, respectively, as measured counterclockwise from the positive x axis. What are the values of (a)
r
⋅
s
and (b) ∣
r
×
s
∣ ? (a) Number Units (b) Number Units
(a) The value of r · s is approximately 15.26 units
(b) The value of |r × s| is approximately 9.47 units.
(a)
The dot product of two vectors is defined as the product of their magnitudes and the cosine of the angle between them.
The formula for dot product is given below:
r · s = |r| |s| cos θ
where
|r| is the magnitude of vector r,
|s| is the magnitude of vector s,
θ is the angle between the vectors.
Using the formula above,
r · s = |r| |s| cos θ
r · s = (3.78)(4.81) cos (50.0° - 317°)
r · s = (3.78)(4.81) cos (-267°)
r · s = (3.78)(4.81)(0.841)
≈ 15.26
Therefore, the value of r · s is approximately 15.26 units.
(b)
The magnitude of the cross product of two vectors is defined as the product of their magnitudes and the sine of the angle between them.
The formula for the magnitude of cross product is given below:
|r × s| = |r| |s| sin θ
where
|r| is the magnitude of vector r,
|s| is the magnitude of vector s,
θ is the angle between the vectors.
Using the formula above,
|r × s| = |r| |s| sin θ
|r × s| = (3.78)(4.81) sin (50.0° - 317°)
|r × s| = (3.78)(4.81) sin (-267°)
|r × s| = (3.78)(4.81)(-0.541)
≈ -9.47
Note that the magnitude of the cross product is always positive, so we take the absolute value of the result.
Therefore, the value of |r × s| is approximately 9.47 units.
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Determine the average irradiation received for the month of December during the hour 10 to 11 am, solar time, on a north-facing collector with tilt 30o and located at Maseru (latitude 29.3o South). The long-term monthly average daily irradiation on a horizontal surface (global radiation) in December is 7.0 kWh/m2.
Do the exercise manually and compare your results with those used by the calculation model given on the link below.
The average irradiation received for the month of December during the hour 10 to 11 am, solar time, on a north-facing collector with tilt 30o and located at Maseru is I = 4.10 kWh/m².
Given :
Latitude of Maseru, φ = 29.3°
Collector tilt angle, β = 30°
Long-term monthly average daily irradiation on a horizontal surface (global radiation) in December, Gm = 7.0 kWh/m²
We know that the average irradiation received on a collector tilted at β with a surface azimuth angle of γ facing true south during the hour 10 to 11 am, solar time can be calculated by using the formula,
I = Gm x [cos (ω) x cos (φ) x cos (β) + sin (φ) x sin (β) x cos (γ - ψ)]
Here, ψ is the difference between the collector's surface azimuth angle and the azimuth angle of the sun, which can be given as, ψ = cos-1 [tan (δ) x cos (λ - λs)] + 180°
where, δ is the declination angle and can be calculated as,
δ = 23.45° x sin [360 x (284 + n) / 365]
Here, n is the day number in the year (1 for January 1st, 365 for December 31st/366 for leap years) and λ is the longitude of the location and λs is the longitude of the standard meridian of the time zone.
For South Africa, the time zone is UTC+2, and thus the standard meridian is 15°E.
Therefore, λs = 15°E + 2.5°E = 17.5°E = 342.5°
The value of δ for December 21st (n = 355) can be calculated as,
δ = 23.45° x sin [360 x (284 + 355) / 365] = -23.14°
The value of ψ can be calculated as,
ψ = cos-1 [tan (-23.14°) x cos (32.56°)] + 180° = 149.37°
Therefore, the average irradiation received for the month of December during the hour 10 to 11 am, solar time, on a north-facing collector with tilt 30° and located at Maseru can be calculated as,
I = 7.0 x [cos (ω) x cos (29.3°) x cos (30°) + sin (29.3°) x sin (30°) x cos (149.37° - ψ)]
For the hour 10 to 11 am, solar time, the value of ω can be calculated as,
ω = 15° x (10.5 - 12) = -22.5°
Therefore, the average irradiation received can be calculated as,
I = 7.0 x [cos (-22.5°) x cos (29.3°) x cos (30°) + sin (29.3°) x sin (30°) x cos (149.37° - 139.3°)] = 4.10 kWh/m²
Thus, the average irradiation received, I = 4.10 kWh/m²
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The position of particular particle as a function of time is given by 9.60t
^
+8.85
^
+1.00t
2
k
^
. Determine the particle's velocity and acceleration as function of time.
The particle's velocity function is given by 9.60 ^ + 2.00t k ^, and its acceleration function is 2.00 k ^.
Given the position function as 9.60t ^ + 8.85 ^ + 1.00t² k ^, we can determine the particle's velocity and acceleration by taking the derivatives of the position function with respect to time.
Velocity function: The velocity of a particle is the derivative of its position with respect to time. Taking the derivative of each component of the position function, we have:
Velocity = (d/dt)(9.60t ^) + (d/dt)(8.85 ^) + (d/dt)(1.00t² k ^).
The derivative of 9.60t with respect to time is 9.60, as it is a constant term. Similarly, the derivative of 8.85 with respect to time is zero, as it is a constant term. The derivative of 1.00t² with respect to time is 2.00t. Therefore, the velocity function can be written as:
Velocity = 9.60 ^ + 2.00t k ^.
Acceleration function: The acceleration of a particle is the derivative of its velocity with respect to time. Taking the derivative of the velocity function, we have:
Acceleration = (d/dt)(9.60 ^) + (d/dt)(2.00t k ^).
The derivative of 9.60 ^ with respect to time is zero, as it is a constant term. The derivative of 2.00t with respect to time is 2.00. Therefore, the acceleration function can be written as:
Acceleration = 2.00 k ^.
To summarize, the particle's velocity function is given by 9.60 ^ + 2.00t k ^, and its acceleration function is 2.00 k ^. The velocity function represents the rate of change of position with respect to time, while the acceleration function represents the rate of change of velocity with respect to time.
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A football is kicked at an angle of 30o with speed of 20 m/s. How far will ball travel before hitting the ground?
a. 15m
b. 25m
c. 35m
d. 45m
How long will the ball stay in the air?
a. 1s
b. 2s
c. 3s
d. 4s
The correct option is (b) 25m.
Time of flight (t) = ?
The time of flight, use the formula Time of flight (t) = 2u sin θ / g= 2 * 20 sin 30 / 9.8≈ 2.04s
Hence, the correct option is (d) 4s.
Given ,Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Horizontal range of projectile (R) = ?
To find the horizontal range of projectile, use the formula Horizontal range of projectile (R) = (u² sin 2θ) / g
Where, g = Acceleration
due to gravity = 9.8 m/s²R = (20² sin 60) / 9.8R
= 25.81
≈ 26 m (approx)
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Consider an isentropic airflow with speed v= 680 m/s, Temperature = 288 K and Pressure= 1.01 N/m².
Determine:
a. Total Temperature
b. Total Pressure
c. Total Density
d. The total enthalpy
e. The flow Mach Number
Given is the velocity of the airflow (v), temperature (T), and pressure (P) as v= 680 m/s, T= 288 K, and P= 1.01 N/m² respectively. We are required to calculate the following quantities:
Total temperatureTotal pressureTotal densityTotal enthalpyFlow Mach numbera.
Total temperatureThe total temperature is given by:Tt = T + (v²/2*Cp)Where, Cp is the specific heat at constant pressureCp = 1005 J/kg-KTt = 288 + ((680²)/(2*1005))= 1001.86 Kb. Total pressureThe total pressure is given by:Pt = P*(1+(γ-1)/2*(Mach)²)^(γ/γ-1)Where, γ is the ratio of the specific heat of the gas= 1.4Mach is the Mach number Pt = 1.01*(1+(1.4-1)/2*(680/343)^2)^(1.4/1.4-1)= 1.287 N/m²c. Total densityThe total density is given by:ρt = P/((R/M)*Tt)Where, R is the specific gas constant= 287 J/kg-KM is the molar mass of the gas= 28.96 gm/molρt = 1.01/((287/28.96)*1001.86)= 0.372 kg/m³d. Total enthalpyThe total enthalpy is given by:Ht = Cp*TtHt = 1005*1001.86= 1,008,300 J/ke. Flow Mach numberThe Mach number is given by:Mach = v/aWhere, a is the speed of sound in the gas. a = √(γ*R*T) = 343 m/sMach = 680/343= 1.98Therefore, the Total Temperature is 1001.86 K.The Total Pressure is 1.287 N/m².The Total Density is 0.372 kg/m³.The Total Enthalpy is 1,008,300 J/ke.The Flow Mach Number is 1.98.
A particle is moving at 4.10 m/s at an angle of 33.5 degrees above the horizontal. Two seconds later, its velocity is 6.05 m/s at an angle of 59.0 degrees below the horizontal. Take the horizontal to be the x-axis. A) What was particles initial velocity vector? B) What was particles final velocity vector? C) What was the magnitude and direction of the particles average acceleration vector during these two seconds?
Initial velocity vector, v1= (4.10 m/s) cos(33.5°) î + (4.10 m/s) sin(33.5°) j= (3.42 î + 2.25 j)
Final velocity vector, v2= (6.05 m/s) cos(59.0°) î - (6.05 m/s) sin(59.0°) j= (3.10 î - 5.23 j)
The magnitude and direction of the particles' average acceleration vector during these two seconds are 3.74 m/s² and -86.4°, respectively.
A particle is moving at 4.10 m/s at an angle of 33.5 degrees above the horizontal. Two seconds later, its velocity is 6.05 m/s at an angle of 59.0 degrees below the horizontal. Take the horizontal to be the x-axis.
A) Initial velocity vector, v = 4.10 m/s and θ = 33.5° above the horizontal.
x-component of the initial velocity vector = v cosθ
y-component of the initial velocity vector = v sinθ
Initial velocity vector, v1= (4.10 m/s) cos(33.5°) î + (4.10 m/s) sin(33.5°) j= (3.42 î + 2.25 j) m/s (rounded to 3 significant figures)
B) Final velocity vector, v = 6.05 m/s and θ = 59.0° below the horizontal.
x-component of the final velocity vector = v cosθ
y-component of the final velocity vector = v sinθ
Final velocity vector, v2= (6.05 m/s) cos(59.0°) î - (6.05 m/s) sin(59.0°) j= (3.10 î - 5.23 j) m/s (rounded to 3 significant figures)
C) Average acceleration vector:
Initial velocity vector, v1= (3.42 î + 2.25 j) m/s
Final velocity vector, v2= (3.10 î - 5.23 j) m/s
Time taken, t = 2 s
Average acceleration, a = (v2 - v1)/t= (3.10 î - 5.23 j - 3.42 î - 2.25 j)/2 s= (-0.16 î - 3.74 j) m/s² (rounded to 3 significant figures)
Magnitude of the average acceleration, |a| = √((-0.16 m/s²)² + (-3.74 m/s²)²)= 3.74 m/s² (rounded to 3 significant figures)
The direction of the average acceleration,θ = tan⁻¹(-3.74/-0.16) = 86.4° (rounded to 1 decimal place)
Therefore, the magnitude and direction of the particles' average acceleration vector during these two seconds are 3.74 m/s² and -86.4°, respectively.
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