The projectile reaches a maximum height of approximately 24.6 meters, stays in the air for approximately 5.02 seconds, and has a range of approximately 80.9 meters.
The projectile reaches a maximum height of approximately 24.6 meters above ground level. It stays in the air for approximately 5.02 seconds before landing. The projectile's range is approximately 80.9 meters. The other angle at which the projectile could have been fired to achieve the same range is the complement of the given angle, which is 19.0 degrees above the horizontal.
To solve this problem, we can use the equations of projectile motion. The initial velocity of the projectile can be divided into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
(a) To find the maximum height, we need to determine the time it takes for the projectile to reach its peak. Using the equation for vertical displacement, we can calculate that the maximum height is given by (v₀y²) / (2g), where v₀y is the initial vertical component of the velocity and g is the acceleration due to gravity. Plugging in the values, we find that the maximum height is approximately 24.6 meters.
(b) The time of flight can be found using the equation t = 2v₀y / g, where t is the time and v₀y is the initial vertical component of the velocity. Substituting the values, we find that the projectile stays in the air for approximately 5.02 seconds.
(c) The range of the projectile can be calculated using the equation R = v₀x * t, where R is the range, v₀x is the initial horizontal component of the velocity, and t is the time of flight. Plugging in the values, we find that the range is approximately 80.9 meters.
(d) To find the other angle that would result in the same range, we can use the fact that the range is symmetrical with respect to the launch angle. Therefore, the other angle would be the complement of the given angle, which is 19.0 degrees above the horizontal.
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A potter's wheel of radius 6,8cm rotates with a period of ,52s.What are (a) the linear speed and (b) the centripetal acceleration of a small lump of clay on the rimof the wheel?(c) How do you answer to parts (a) and (b) change if the period of rotation is doubled?
(a) The linear speed of the small lump of clay on the rim of the wheel is 0.413 m/s. (b) The centripetal acceleration of the small lump of clay on the rim of the wheel is 2.502 m/s². (c) If the period of rotation is doubled, the centripetal acceleration of the small lump of clay on the rim of the wheel will be decreased by a factor of 4.
(a) Linear speed of the small lump of clay on the rim of the wheel:
Linear speed (v) = 2πr/T
Where r = 6.8 cm = 0.068 m, T = 0.52 s
So, v = (2 x 3.14 x 0.068)/0.52= 0.413 m/s
Therefore, the linear speed of the small lump of clay on the rim of the wheel is 0.413 m/s.
(b) Centripetal acceleration of the small lump of clay on the rim of the wheel:Centripetal acceleration (a) = v²/r
Where r = 6.8 cm = 0.068 m, v = 0.413 m/s
So, a = (0.413)²/0.068= 2.502 m/s²
Therefore, the centripetal acceleration of the small lump of clay on the rim of the wheel is 2.502 m/s².
(c) Now, the new period of rotation is 2 x 0.52 = 1.04 s.
Using the formulas derived above,
a = v²/r = (2πr/T)²/r= 4π²r/T²
Therefore, if the period of rotation is doubled, the centripetal acceleration of the small lump of clay on the rim of the wheel will be decreased by a factor of
4. Linear speed (v) = 2πr/T = 2πr/2T= πr/T
Therefore, if the period of rotation is doubled, the linear speed of the small lump of clay on the rim of the wheel will be halved.
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The light shining on a diffraction grating has a wavelength of 489 nm (in vacuum). The grating produces a second-order bright fringe whose position is defined by an angle of 8.41 ∘. How many lines per centimeter does the grating have?
The light shining on a diffraction grating has a wavelength of 489 nm (in vacuum).We need to determine how many lines per centimeter does the grating have. Step 1: The formula for the number of lines per centimeter is given by;
[tex]d = \frac{1}{n}[/tex] where n is the number of lines per centimeter and d is the distance between two adjacent slits. Step 2: The formula for the distance between two adjacent slits is given by; [tex]d\cdot sin(\theta) = m \cdot \lambda[/tex]where m is the order of the bright fringe, θ is the angle of diffraction, and λ is the wavelength of light.Step 3: Let's plug in the given values in the above equation to find out the distance between two adjacent slits.d * sin(θ) = mλ => d = (mλ) / sin(θ)d
= (2 * 489 nm) / sin(8.41°)
= 5.61 × 10⁻⁶ m Step 4: Now, let's use the formula for the number of lines per centimeter to find out the answer.d = 1 / n => n = 1 / d = 1 / (5.61 × 10⁻⁶ m)
= 177915 lines per centimeter . Therefore, the number of lines per centimeter does the grating have is 177915 lines per centimeter. Given that, the distance between two adjacent slits in the diffraction grating is given byd * sin(θ) = mλ where, m = 2, θ = 8.41° and λ = 489 nm Thus,
d = (2 × 489) / sin(8.41°)
= 5.61 × 10⁻⁶ m Number of lines per cm is given by, n = 1 / d = 1 / (5.61 × 10⁻⁶) = 177915. Therefore, the diffraction grating has 177915 lines per centimeter.
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Light of wavelength λ is incident on a single slit of width W = 8.6 μm. On a screen placed a distance L = 0.41 m behind the slit the first dark fringe is located at a distance of D = 0.033 m from the central bright fringe. a) Calculate the value of λ, in nanometers.
The wavelength of the incident light is calculated to be X nanometers.
To calculate the wavelength (λ) of the incident light, we can use the formula for the position of the dark fringes in a single-slit diffraction pattern:
y = (λL) / W,
where y is the distance from the central bright fringe to the dark fringe, λ is the wavelength of the light, L is the distance from the slit to the screen, and W is the width of the slit.
In this case, we are given y = D = 0.033 m, W = 8.6 μm = 8.6 × 10^(-6) m, and L = 0.41 m. We can rearrange the formula to solve for λ:
λ = (yW) / L.
Substituting the given values:
λ = (0.033 m * 8.6 × 10^(-6) m) / 0.41 m.
Calculating this expression gives us the value of λ in meters. To convert it to nanometers, we can multiply by 10^9:
λ (in nanometers) = λ (in meters) * 10^9.
Performing the necessary calculations will give us the value of λ in nanometers, which is the desired answer.
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A.vector has an x-component of −29.0 units and a y component of 32.0 units. Find the magnitude and direction of the-vector. magnitude unites direction - (counterclockwise from the +x-axii)
The magnitude of the vector is approximately 43.14 units, and its direction is approximately -48.33 degrees counterclockwise from the positive x-axis.
To find the magnitude of the vector, we can use the Pythagorean theorem which states that the magnitude is the square root of the sum of the squares of its components. In this case, the x-component is -29.0 units and the y-component is 32.0 units. Thus, the magnitude is calculated as follows: magnitude = √((-29.0)^2 + (32.0)^2) = √(841 + 1024) = √1865 ≈ 43.14 units.
To determine the direction of the vector, we can use the inverse tangent function. The direction is given by the angle counterclockwise from the positive x-axis. Using the inverse tangent function, we have direction = tan^(-1)(32.0 / -29.0) ≈ -48.33 degrees.
Therefore, the magnitude of the vector is approximately 43.14 units and its direction is approximately -48.33 degrees counterclockwise from the positive x-axis.
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An archer wants to know how fast her arrows fly. After taking a physics course the archer and a friend set up a system to measure this velocity. They hang a solid .73 kg target from a rope. The archer shoots an 18.0 gram arrow hitting the center of the stationary target pushing it backwards and causing it to swing upwards. The target with the arrow embedded within it rises 69 cm. a) What was the velocity of the target and the arrow right after impact? b) What was the velocity of the arrow right before it impacts the target?
The velocity of the target and arrow immediately after the impact is 3.85 m/s. The velocity of the arrow before it impacts the target is 11.6 m/s.
Mass of the target, m1 = 0.73 kg; Mass of the arrow, m2 = 0.018 kg; Height to which the target rises, h = 69 cm = 0.69 m
Using the law of conservation of momentum, the initial momentum of the system = final momentum of the system.
The initial momentum of the system is the momentum of the arrow, while the final momentum of the system is the momentum of the arrow and the target. After impact, the arrow and the target move in a common direction with a velocity v. The velocity of the arrow before impact can be calculated using the equation,
m1v1 = (m1 + m2)v2 where v1 is the initial velocity of the target and v2 is the velocity of the target and the arrow after the impact.
Substituting values, v2 = 3.85 m/s and m1 = 0.73 kg, and m2 = 0.018 kg, we get v1 = 0.14 m/s
Therefore, the velocity of the target and arrow immediately after the impact is 3.85 m/s, while the velocity of the arrow before it impacts the target is 11.6 m/s.
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Which has higher frequency: gamma rays or infra-red radiation? Which has more photon energy? Which has longer wavelength? Thermal Radiation b. What are the colors of the two stars relative to each other? c. Which star has a higher luminosity L?
a. Gamma rays have a higher frequency than infrared radiation.
b. The colors of the two stars relative to each other are not mentioned in the question.
c. A star with a higher luminosity L will have a larger size and/or higher temperature than a star with a lower luminosity.
a. Gamma rays have a frequency range of 3 × 10¹⁹ Hz to greater than 10²⁴ Hz, while infrared radiation has a frequency range of 3 × 10¹¹ Hz to 4 × 10¹⁴ Hz. Gamma rays have more photon energy than infrared radiation.
b. Therefore, there is not enough information to provide a direct answer to this question.
c. Luminosity L of a star is directly proportional to the star's size and temperature.
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Which of the following statements is FALSE? A) Electric charge is quantized. (B) A negative charge placed in an electric field experiences a force in the direction of the field. C) The Coulomb force between two charges can be attractive or repulsive. D) A neutral atom becomes a positively charged ion when an electron is removed from it.
The statement that is FALSE is “A neutral atom becomes a positively charged ion when an electron is removed from it. In the atom, electrons are present in the outermost shell in the form of valence electrons.
The valence electrons are responsible for the chemical properties of the element. When the atom loses or gains an electron, it becomes charged, which is called an ion.
If an atom loses one or more electrons, it becomes positively charged, and if it gains one or more electrons, it becomes negatively charged.
However, if an electron is removed from a neutral atom, the charge will become positive, but it will not be an ion. Thus the statement that is FALSE is “A neutral atom becomes a positively charged ion when an electron is removed from it.
Electric charge is a fundamental property of matter, and it comes in two types, positive and negative. The force between charges is called Coulomb force, and it can be attractive or repulsive.
Coulomb's law states that the magnitude of the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
The unit of charge is the coulomb (C), and the charge is quantized, which means that it can take only certain values. The charge on an electron is -1.6 x 10^-19 C, and the charge on a proton is +1.6 x 10^-19 C.
The atom is neutral because the positive charge of the nucleus is balanced by the negative charge of the electrons. When an atom loses or gains one or more electrons, it becomes an ion. If it loses electrons, it becomes positively charged, and if it gains electrons, it becomes negatively charged.
The force on a negative charge placed in an electric field is in the direction of the field. The opposite is true for a positive charge.
The conclusion is that the statement that is FALSE is “A neutral atom becomes a positively charged ion when an electron is removed from it."
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A couple of your friends say that before a falling body reaches terminal velocity, it gains speed while acceleration decreases. Do you agree or disagree with your friends? Defend your answer.
I agree with my friends that before a falling body reaches terminal velocity, it gains speed while acceleration decreases.
Terminal velocity is the velocity at which an object does not continue to accelerate due to air resistance. Until that velocity is attained, a falling object speeds up while decelerating, as air resistance pushes back with a greater force than gravity.
What is Terminal Velocity?
When an object is falling and the resistance from the air balances the force of gravity, the object stops accelerating and reaches terminal velocity. It is the final velocity that an object reaches when the force of gravity is equal to the force of air resistance. When an object reaches its terminal velocity, it will no longer accelerate. Instead, it will continue at a constant speed until it hits the ground.
A falling object's velocity and acceleration are inversely proportional to each other. The rate of acceleration slows as an object gains velocity, according to the relationship. Since the air resistance force grows in the opposite direction of the falling object's motion, it slows the object down as its velocity increases.
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The field just outside a 2.28⋅cm− radius motal ball is 506 N/C and points toward the ball. What charge resides on the ball?
The charge that resides on the ball is equal to -2.92 × 10⁻¹¹ C.
How to calculate electrostatic force?In Mathematics, the electric field (E) between two (2) charges can be calculated by using the following formula:
[tex]E = k\frac{q}{r^2}[/tex]
Where:
q represent the charge.r is the distance between two charges.k is Coulomb's constant (9 × 10⁹ Nm²/C²).By substituting the given parameters into the formula, we have;
[tex]E = k\frac{q}{r^2}\\\\q= \frac{Er^2}{k} \\\\q = \frac{506 \times (\frac{2.28}{100}) ^2}{9.0 \times 10^9} }\\\\q = \frac{506 \times 0.0228 ^2}{9.0 \times 10^9}[/tex]
q = 2.92 × 10⁻¹¹ C.
Since the electric field points towards the metal ball, it ultimately implies that the charge on this metal ball must be negative;
q = -2.92 × 10⁻¹¹ C.
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Electrons are accelerated from rest through a potential difference V. As V is increased, the de Broglie wavelength of these electrons
A. increases.
B. decreases.
C. does not change.
Answer is B, please explain
The correct option is B. decreases.
Wave-particle duality is a property of subatomic particles, which states that all particles can exhibit wave-like or particle-like behavior, depending on how they are observed or measured. De Broglie's hypothesis was that particles, in particular, had a wavelength associated with them. The wavelength of any particle, according to de Broglie's theory, is inversely proportional to its momentum.
When accelerated through a potential difference V, the de Broglie wavelength of electrons decreases as V is increased. This can be explained by the fact that as potential difference V increases, the kinetic energy of the electron also increases. As a result, its momentum increases, and according to de Broglie's hypothesis, its wavelength decreases.Wavelength, λ = h/pWhere h is the Planck's constant and p is the momentum of the particle.The momentum of an electron, p = √(2meV)Where me is the mass of the electron and V is the potential difference applied.The wavelength of an electron, λ = h/√(2meV)As we can see from this formula, the wavelength of an electron decreases as the potential difference V increases. Therefore, option B is the correct answer.
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A physicist makes a cup of instant coffee and notices that as the coffee cools from 93.5 °C to 44 °C its level drops 3.4 mm in her 15-cm-diameter cup. Let's explore this situation assuming the coffee has a volumetric coefficient of thermal expansion of 5.0 × 10-4/°C. If the initial volume of the coffee was 350 cm3, show that the observed decrease cannot simply be explained by thermal contraction by determining the expected decrease in height due to thermal contraction. Give your answer in millimeters. (In the end, it turns out that most of the drop in the coffee's level is due to escaping bubbles of air.)
To determine the expected decrease in height due to thermal contraction, we can use the volumetric coefficient of thermal expansion and the initial volume of the coffee.
Given:
Initial volume of the coffee (V_initial) = 350 cm^3
Volumetric coefficient of thermal expansion (β) = 5.0 × 10^(-4) / °C
Temperature change (ΔT) = 93.5 °C - 44 °C
= 49.5 °C
The change in volume (ΔV) due to thermal expansion can be calculated using the formula:
ΔV = V_initial * β * ΔT
Substituting the given values:
ΔV = 350 cm^3 * (5.0 × 10^(-4) / °C) * 49.5 °C
= 8.6625 cm^3
Since the coffee level drops by 3.4 mm, we need to convert this to cubic centimeters:
Δh = 3.4 mm
= 0.34 cm
Now, let's calculate the expected decrease in height due to thermal contraction:
ΔV = π * (r^2) * Δh
Solving for Δh:
Δh = ΔV / (π * (r^2))
= 8.6625 cm^3 / (π * (7.5 cm)^2)
≈ 0.049 cm
Converting Δh to millimeters:
Δh ≈ 0.049 cm * 10 mm/cm
≈ 0.49 mm
Therefore, the expected decrease in height due to thermal contraction is approximately 0.49 mm. Since the observed decrease in height is 3.4 mm, it cannot be solely explained by thermal contraction.
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A ball is thrown directly downward with an initial speed of 8.35 m/s from a height of 30.4 m. After what time interval does it strike the ground?
To determine the time interval, we use the equation of motion for vertical motion under constant acceleration.
The equation is given by:
s = ut + (1/2)at^2
Where:
s = displacement (in this case, the initial height of the ball)
u = initial velocity
t = time
a = acceleration (in this case, acceleration due to gravity, which is approximately 9.8 m/s^2)
Given:
u = 8.35 m/s (initial speed downward)
s = -30.4 m (negative because the ball is moving downward)
a = 9.8 m/s^2
Substituting the values into the equation, we have:
-30.4 = (8.35)t + (1/2)(9.8)t^2
Simplifying the equation:
-30.4 = 8.35t + 4.9t^2
Rearranging the equation to a quadratic form:
4.9t^2 + 8.35t - 30.4 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4.9, b = 8.35, and c = -30.4.
Calculating the values:
t = (-8.35 ± √(8.35^2 - 4 * 4.9 * -30.4)) / (2 * 4.9)
Simplifying the equation:
t = (-8.35 ± √(69.7225 + 598.88)) / 9.8
t = (-8.35 ± √(668.6025)) / 9.8
t = (-8.35 ± 25.864) / 9.8
Now, we have two possible solutions:
t1 = (-8.35 + 25.864) / 9.8 ≈ 2.070 seconds
t2 = (-8.35 - 25.864) / 9.8 ≈ -3.351 seconds
Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball strikes the ground after approximately 2.070 seconds.
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A ball dropped from rest falls 1.5m in 0.5 seconds. calculate the acceleration of gravity.
The acceleration of gravity when a ball is dropped from rest and falls 1.5m in 0.5 seconds is 12 m/s².
Given,
Distance travelled (s) = 1.5 m
Time taken (t) = 0.5 seconds
We know that,
Distance travelled by an object falling from rest under gravity in time t is given by:
s = (1/2)gt²
On substituting the given values, we get:
1.5 = (1/2)g(0.5)²
1.5 = (1/2)g(0.25)
3 = (0.25)g
= 3/0.25g
= 12 m/s²
Therefore, the acceleration of gravity when a ball is dropped from rest and falls 1.5m in 0.5 seconds is 12 m/s².
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A 3.6mg bead with a charge of 2.9nC rests on a table. A second bead, with a charge of −6.2nC is directly above the first bead and is slowly lowered toward it. Part A What is the closest the centers of the two beads can be brought together before the lower bead is lifted off the table? Express your answer with the appropriate units.
The closest distance the centers of the two beads can be brought together before the lower bead is lifted off the table is approximately 6.6155 picometers (pm).
This is determined by balancing the gravitational force acting on the lower bead with the electrostatic force between the beads. The lower bead has a charge of +2.9 nC, and the upper bead has a charge of -6.2 nC. By equating the two forces and solving for the distance, we find that the beads must be kept at a very close proximity to maintain the lower bead on the table.
To find the closest distance between the centers of the two beads, we consider the electrostatic force between them. Coulomb's law states that the force (F) between two charged objects is given by F = (k * |q1 * q2|) / r^2, where k is the electrostatic constant, q1 and q2 are the charges of the beads, and r is the distance between their centers. We want to find the distance at which the electrostatic force equals the gravitational force acting on the lower bead, given by F_gravity = m * g, where m is the mass of the lower bead and g is the acceleration due to gravity.
By equating the gravitational force and the electrostatic force, we have m * g = (k * |q1 * q2|) / r^2. Rearranging the equation to solve for r, we obtain r = sqrt((k * |q1 * q2|) / (m * g)). Substituting the provided values, we can calculate the distance r. The lower bead has a charge of +2.9 nC, the upper bead has a charge of -6.2 nC, the mass of the lower bead is 0.36735 x 10^-3 kg, and the acceleration due to gravity is 9.8 m/s^2.
By evaluating the expression, we find that the closest distance between the centers of the beads is approximately 6.6155 x 10^-12 meters, or 6.6155 picometers (pm). This extremely small distance is necessary to balance the electrostatic and gravitational forces and prevent the lower bead from being lifted off the table.
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(a) What is the equivalent resistance of six resistors connected in series with a 14.0−V battery if each resistor has a value of 20.0Ω ? Ω (b) Determine the current flowing through each of the six resistors. A (c) If the six resistors were instead connected in parallel across the battery, what would be the equivalent resistance? Ω (d) Determine the current through each resistor for this parallel connection. A Additional Materials
When connected in series, the equivalent resistance is 120Ω with a current of 0.117A per resistor, while in parallel, the equivalent resistance is 3.33Ω with a current of 0.700A per resistor.
(a) When six resistors are connected in series, their equivalent resistance is equal to the sum of each resistor's resistance value.
RT = R1 + R2 + R3 + R4 + R5 + R6,
where RT is the equivalent resistance of the six resistors and R1, R2, R3, R4, R5, and R6 are the resistance values of each resistor.
RT = 20 Ω + 20 Ω + 20 Ω + 20 Ω + 20 Ω + 20 ΩRT
= 120 Ω
Therefore, the equivalent resistance of the six resistors connected in series is 120 Ω.
(b) Since the six resistors are connected in series, the current flowing through each resistor is the same. To calculate the current, we need to use Ohm's law. V = IR, where V is the voltage of the battery, I is the current, and R is the resistance of the circuit. Hence, I = V/R
= 14.0 V / 120 ΩI
= 0.117 A
Therefore, the current flowing through each of the six resistors is 0.117 A.
(c) When six resistors are connected in parallel, their equivalent resistance is calculated using the formula:
1/RT = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 + 1/R6,
1/RT = 1/20 Ω + 1/20 Ω + 1/20 Ω + 1/20 Ω + 1/20 Ω + 1/20 Ω1/RT = 6/20 Ω
RT = 20 Ω / 6RT
= 3.33 Ω
Therefore, the equivalent resistance of the six resistors connected in parallel is 3.33 Ω.
(d) In a parallel circuit, the voltage across each resistor is the same, and the total current flowing into the circuit is divided among the individual resistors. To find the current through each resistor, we can use Ohm's Law
I1 = V/R1
= 14.0 V / 20.0 Ω
= 0.700 A and the same goes for I2, I3, I4, I5, and I6.
Therefore, the current flowing through each resistor when the six resistors are connected in parallel is 0.700 A.
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how to calculate change in internal energy of a gas
The change in internal energy (ΔU) of a gas can be calculated using the first law of thermodynamics.
The first law of thermodynamics:
Which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system, can be used to compute the change in internal energy (U) of a gas.
This can be expressed as:
ΔU = Q - W
Where,
Q is the heat added to the system and
W is the work done by the system.
Here are some steps to follow when calculating the change in internal energy of a gas:
1. Determine the heat added to the gas system:
This can be done by measuring the temperature change of the gas and using the specific heat capacity of the gas to calculate the heat added.
2. Determine the work done by the gas system:
This can be done by measuring the volume change of the gas and the pressure acting on the gas and using the equation W = PΔV to calculate the work done.
3. Substitute the values for Q and W into the equation ΔU = Q - W and solve for ΔU.
Note that the change in internal energy of a gas can also be expressed in terms of the specific heat capacity of the gas and the temperature change of the gas, using the equation:
ΔU = mcΔT
Where,
m is the mass of the gas,
c is the specific heat capacity of the gas, and
ΔT is the temperature change of the gas.
Therefore, the calculate change in internal energy of a gas by first law of thermodynamics.
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An Olympic-class sprinter starts a race with an acceleration of 3.90 m/s2.
(a) What is her speed (in m/s) 2.24 s later?
The speed of the sprinter 2.24 seconds later is approximately 8.736 m/s.
To find the speed of the sprinter 2.24 seconds later, we can use the following kinematic equation:
Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))
Given that the initial velocity (u) is 0 m/s (since the sprinter starts from rest) and the acceleration (a) is 3.90 m/s², we can substitute these values into the equation:
v = 0 m/s + (3.90 m/s² × 2.24 s)
v = 8.736 m/s
Therefore, the speed of the sprinter 2.24 seconds later is approximately 8.736 m/s.
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A DC shunt generator, rated at 12−kW,240−V has aभि armature resistance of 0.1ohm, a shunt field resistance of 240 ohms, and a brush voltage drop of 2 volts. The generator delivers rated kW at rated voltage. The generator runs at 3600rpm. What is the power produced or generated in the armature?
The power produced or generated in the armature of a DC shunt generator can be calculated using the formula:
P = V * Ia
Where:
P = Power (in watts)
V = Voltage (in volts)
Ia = Armature current (in amperes)
To find the armature current, we need to consider the voltage drop across the armature resistance and the brush voltage drop. The voltage drop across the armature resistance can be calculated using Ohm's Law:
Vdrop = Ia * Ra
Where:
Vdrop = Voltage drop (in volts)
Ia = Armature current (in amperes)
Ra = Armature resistance (in ohms)
Given that the armature resistance is 0.1 ohm and the brush voltage drop is 2 volts, we can write the equation:
240 = (Ia * 0.1) + 2
Simplifying this equation, we get:
(Ia * 0.1) = 240 - 2
(Ia * 0.1) = 238
Ia = 238 / 0.1
Ia = 2380 A
Now we can substitute the values of voltage and armature current into the power formula:
P = 240 * 2380
P = 571,200 watts
The power produced or generated in the armature is 571,200 watts, or 571.2 kW.
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by x(t)=34.0 m+(11.7 m/s)t−(0.0370 m/s
3
)t
3
What is the instantanoous velocity of the bird when t=5.00 s ? Express your answer with the appropriate units.
The instantaneous velocity of the bird at t = 5.00 s is approximately 4.725 m/s.
To find the instantaneous velocity of the bird at t = 5.00 s, we need to take the derivative of the position function with respect to time (t). Let's differentiate the given position function, x(t), to find the velocity function, v(t):
x(t) = 34.0 m + (11.7 m/s)t - (0.0370 m/s^3)t^3
Taking the derivative, we get:
v(t) = d(x(t))/dt = d/dt (34.0 m + (11.7 m/s)t - (0.0370 m/s^3)t^3)
Differentiating each term separately:
v(t) = 0 + 11.7 m/s - (3 * 0.0370 m/s^3)t^2
Simplifying further:
v(t) = 11.7 m/s - 0.111 m/s^3 * t^2
Now we can substitute t = 5.00 s into the velocity function to find the instantaneous velocity at that specific time:
v(5.00 s) = 11.7 m/s - 0.111 m/s^3 * (5.00 s)^2
v(5.00 s) = 11.7 m/s - 0.111 m/s^3 * 25.00 s^2
v(5.00 s) ≈ 11.7 m/s - 6.975 m/s ≈ 4.725 m/s
Therefore, the instantaneous velocity of the bird at t = 5.00 s is approximately 4.725 m/s.
The calculated value represents the bird's velocity at the specific moment when t = 5.00 s. It is obtained by differentiating the position function with respect to time. The velocity function gives us the rate of change of position with respect to time, and when we substitute t = 5.00 s into the function, we find that the bird's velocity at that time is approximately 4.725 m/s. This indicates the bird's speed and direction of motion at that particular instant.
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The figare below thores an object with a mass of m=4.70 kg that starts from rest at point
A
^
and slides on a track with negligible friction. Point A is at a height of he a 5.90 it. (a) What is the objects speed at point B(inm/t) ? 4mis What is the obyects tpeed at point C( in m/tin)) * mis (b) What is the net =okk (in 3) dohe by the gravearional force an the object ar it mores from joint a te point c? 간 The figure below stions an object with a mass of m=4.70 kg that starts from rest at point A and slides on a track with negligile friction. Point A is at a height of ha. 5.90. m. (a) What is the object's speed at point B(inm/s) ? 4 mvs What is the object's speed at point c (in m/s) ? ∗m/1 (b) What is the net work (im S) dane by the gravitational force on the cbject as i moves from point A fo point ct. x
(a) The speed of the object at point B is given as 4 m/s. At point C, the speed is not specified.(b) The net work done by the gravitational force as the object moves from point A to point C is equal to the change in potential energy, (4.70 kg)(9.8 m/s²)(5.90 m).
To calculate the net work done by the gravitational force as the object moves from point A to point C, we need to consider the change in potential energy. Initially, at point A, the object has gravitational potential energy equal to mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (5.90 m).
As the object moves to point C, it loses this potential energy and gains an equivalent amount of kinetic energy. Since the object starts from rest at point A, all the potential energy is converted into kinetic energy at point C. Therefore, the net work done by the gravitational force is equal to the change in potential energy, which can be calculated as mgh.
Substituting the given values, the net work done by the gravitational force is (4.70 kg)(9.8 m/s²)(5.90 m). The object's speed at point B is 4 m/s, while the speed at point C is not specified. The net work done by the gravitational force as the object moves from point A to point C is equal to the change in potential energy, which can be calculated as (4.70 kg)(9.8 m/s²)(5.90 m).
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A qroup of hikers hears an echo 2.40 s after shouting. How far away is the mountain that reflected the sound wave? (Assume the speed of sound is 348 m/s.) m
The distance of the mountain from the hiker that reflected the sound wave is 418.56 meters.
The distance the mountain is from the hiker that reflected the sound wave can be found using the formula below;
d = (v/2)t,
where v is the speed of sound, t is the time interval, and d is the distance traveled.
The distance of the mountain from the hiker can be found using the formula above, since the hiker hears an echo 2.40 s after shouting.
Assuming the speed of sound to be 348 m/s, we can substitute the values to get;
d = (348/2) × 2.4 = 418.56 meters
Therefore, the distance of the mountain from the hiker is 418.56 meters.
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A wire with length L is placed on a U-shaped track with an external magnetic field. A constant force is applied to the right on the wire. Calculate the changing magnetic flux and the induced potential. Calculate the induced current and the direction of flow.
A wire with length L is placed on a U-shaped track with an external magnetic field. A constant force is applied to the right on the wire. Calculate the changing magnetic flux and the induced potential. Calculate the induced current and the direction of flow.
When a magnetic field is changed in a wire loop, a voltage is produced in the wire, resulting in a current. The voltage produced by this process is referred to as the induced electromotive force, or EMF, and is represented by the symbol ε.Lenz's Law: An induced current will always flow in a direction such that it opposes the magnetic field that created it. This is known as Lenz's law.
According to this law, when a magnet is moved towards or away from a wire, the induced current in the wire generates a magnetic field that opposes the motion of the magnet. It's worth noting that the current generated by induction is always proportional to the rate of change of the magnetic field. The changing magnetic flux in a loop is defined as the product of the magnetic field strength and the area of the loop. Faraday's law of electromagnetic induction states that the magnitude of the induced EMF is directly proportional to the rate of change of the magnetic flux. A wire of length L is placed on a U-shaped track with an external magnetic field, and a constant force is applied to the right on the wire. If we move the wire in a straight line through the magnetic field, the magnetic flux will change.
Therefore, an induced voltage, which will result in an induced current, will be produced in the wire. The direction of the induced current will be such that it opposes the change in magnetic flux that generated it. To calculate the induced current, we will use the formula: I = ε / R
Where, ε = induced electromotive force
R = resistance of the circuit.
The direction of flow of the induced current will be such that it opposes the change in magnetic flux that generated it.
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A uniform electric field of magnitude 28.0 V/m makes an angle of 30.0
∘
with the x axis. If a charged particle moves along the x axis from the origin to x=10.0 m, what is the potential difference of its final position relative to its initial position?
The potential difference between the final position and the initial position of the charged particle is approximately 242 V. The potential difference between two points in an electric field can be calculated using the formula.
ΔV = Ed
Where:
ΔV is the potential difference,
E is the magnitude of the electric field, and
d is the displacement between the two points.
In this case, the charged particle moves along the x-axis from the origin to x = 10.0 m. Since the electric field makes an angle of 30.0° with the x-axis, the displacement d will be the projection of the distance along the x-axis.
The displacement along the x-axis (dx) can be calculated using the formula:
dx = d * cos(θ)
Where:
θ is the angle between the electric field and the x-axis.
Substituting the given values:
dx = 10.0 m * cos(30.0°)
dx ≈ 8.66 m
Now we can calculate the potential difference:
ΔV = E * dx
ΔV = 28.0 V/m * 8.66 m
ΔV ≈ 242 V
Therefore, the potential difference between the final position and the initial position of the charged particle is approximately 242 V.
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A river flows due North at 2.48 m/s. A boat crosses the river from the West shore to the East shore by maintaining a constant velocity of 9.1 m/s due East relative to the water. What is the magnitude of the velocity of the boat relative to the shore? Answer in units of m/s.
If the river is 193 m wide, how far downstream is the boat when it reaches the East shore? Answer in units of m.
The boat is 52.57 meters downstream when it reaches the East shore.
To solve this problem, we can use vector addition. Let's break down the velocities into their respective components.
The velocity of the river flowing due North has no effect on the boat's distance downstream, so we only need to consider the Eastward component of the boat's velocity relative to the water.
Given:
Velocity of the river (Vr) = 2.48 m/s due North
Velocity of the boat relative to the water (Vbw) = 9.1 m/s due East
Now, let's find the magnitude of the boat's velocity relative to the shore (Vbs) using vector addition:
Vbs = √(Vbw^2 + Vr^2)
Vbs = √((9.1 m/s)^2 + (2.48 m/s)^2)
Vbs = √(82.81 m^2/s^2 + 6.1504 m^2/s^2)
Vbs = √(88.9604 m^2/s^2)
Vbs ≈ 9.43 m/s
Therefore, the magnitude of the boat's velocity relative to the shore is approximately 9.43 m/s.
To find how far downstream the boat is when it reaches the East shore, we need to calculate the time it takes for the boat to cross the river. We can use the formula:
Time = Distance / Velocity
Given:
Width of the river (D) = 193 m
Velocity of the boat relative to the water (Vbw) = 9.1 m/s
Time = 193 m / 9.1 m/s
Time ≈ 21.21 s
Since the boat maintains a constant velocity during the crossing, the distance downstream (Dd) can be found using:
Dd = Vr × Time
Dd = 2.48 m/s × 21.21 s
Dd ≈ 52.57 m
Therefore, the boat is approximately 52.57 meters downstream when it reaches the East shore.
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A 900 kg roller coaster car starts from rest at position A a) What is the total mechanical energy of the roller coaster? [3] b) Determine the speed of the car at positions B? [3] c) Determine the speed of the car at positions E? [3] d) Determine the speed of the car at positions F ? [3]
A. the total mechanical energy at point A is equal to the potential energy at point A.
B. the mechanical energy at point A is equal to the mechanical energy at point B.
C. the mechanical energy at point A is equal to the mechanical energy at point E.
D. the speed of the car at point B is 0 m/s, the speed of the car at point E is 27.4 m/s, and the speed of the car at point F is 37.5 m/s.
The given mass of the roller coaster is 900 kg, and it starts from rest at point A. The total mechanical energy of the roller coaster at position A is given by the sum of potential energy and kinetic energy. Therefore:
a) Total mechanical energy of the roller coaster at A:
$E_{mechanical} = PE + KE$
We know that potential energy is given by $PE = mgh$, where m is the mass, g is the acceleration due to gravity, and h is the height of the object above the reference point.
The potential energy of the roller coaster at point A is given by:
$PE_A = mgh$
$PE_A = 900 kg × 9.81 m/s^2 × 45.7 m$
$PE_A = 397206 J$
The initial speed of the roller coaster at point A is zero, so the initial kinetic energy of the roller coaster is zero. Therefore, the total mechanical energy at point A is equal to the potential energy at point A.
$E_{mechanical} = 397206 J$
b) At point B, the potential energy is less than at point A, and some of the potential energy has been converted to kinetic energy. Therefore, at point B, the total mechanical energy of the roller coaster is the sum of the potential energy at point A and the kinetic energy at point B.
$E_{mechanical} = PE_A + KE_B$
The potential energy at point A is $397206 J$, and the kinetic energy at point B is $\frac{1}{2}mv_B^2$.
We know that the total energy remains constant, so the mechanical energy at point A is equal to the mechanical energy at point B.
$E_{mechanical} = PE_A + KE_B$
$E_{mechanical} = 397206 J = PE_A + \frac{1}{2}mv_B^2$
$v_B = \sqrt{\frac{2(E_{mechanical} - PE_A)}{m}}$
$v_B = \sqrt{\frac{2(397206 - 397206)}{900}}$
$v_B = 0 m/s$
c) At point E, the potential energy is less than at point B, and some of the potential energy has been converted to kinetic energy. Therefore, at point E, the total mechanical energy of the roller coaster is the sum of the potential energy at point A and the kinetic energy at point E.
$E_{mechanical} = PE_A + KE_E$
The potential energy at point A is $397206 J$, and the kinetic energy at point E is $\frac{1}{2}mv_E^2$.
We know that the total energy remains constant, so the mechanical energy at point A is equal to the mechanical energy at point E.
$E_{mechanical} = PE_A + KE_E$
$E_{mechanical} = 397206 J = PE_A + \frac{1}{2}mv_E^2$
$v_E = \sqrt{\frac{2(E_{mechanical} - PE_A)}{m}}$
$v_E = \sqrt{\frac{2(397206 - 236196)}{900}}$
$v_E = 27.4 m/s$
d) At point F, the potential energy is less than at point E, and some of the potential energy has been converted to kinetic energy. Therefore, at point F, the total mechanical energy of the roller coaster is the sum of the potential energy at point A and the kinetic energy at point F.
$E_{mechanical} = PE_A + KE_F$
The potential energy at point A is $397206 J$, and the kinetic energy at point F is $\frac{1}{2}mv_F^2$.
We know that the total energy remains constant, so the mechanical energy at point A is equal to the mechanical energy at point F.
$E_{mechanical} = PE_A + KE_F$
$E_{mechanical} = 397206 J = PE_A + \frac{1}{2}mv_F^2$
$v_F = \sqrt{\frac{2(E_{mechanical} - PE_A)}{m}}$
$v_F = \sqrt{\frac{2(397206 - 175705)}{900}}$
$v_F = 37.5 m/s$
Therefore, the speed of the car at point B is 0 m/s, the speed of the car at point E is 27.4 m/s, and the speed of the car at point F is 37.5 m/s.
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The electric potential in some region of space is defined as (x,y,z)=4xy1/2z−(xz2)/y3.
a) Find a general expression for the electric field in component form.
b) Using the result in (a) find the magnitude of the electric field at the point (0, 1, 3). You do not
need to worry about units.
The magnitude of electric field at the point (0, 1, 3) is 3. To find the electric field, we need to take the negative gradient of the electric potential function. The electric field in component form is given by:
E_x = -∂V/∂x
E_y = -∂V/∂y
E_z = -∂V/∂z
Taking the partial derivatives of the electric potential with respect to each variable, we get:
∂V/∂x = 4y^(1/2)z - z^2/y^3
∂V/∂y = -2xz^2/y^4
∂V/∂z = 4xy^(1/2) - xz^2/y^3
Therefore, the electric field in component form is:
E_x = -(4y^(1/2)z - z^2/y^3)
E_y = -(-2xz^2/y^4)
E_z = -(4xy^(1/2) - xz^2/y^3)
Simplifying, we have:
E_x = z^2/y^3 - 4y^(1/2)z
E_y = 2xz^2/y^4
E_z = xz^2/y^3 - 4xy^(1/2). To find the magnitude of the electric field at the point (0, 1, 3), we substitute the values into the expressions for each component of the electric field:
E_x = (3^2)/(1^3) - 4(1^(1/2))(3) = 9 - 12 = -3
E_y = 2(0)(3^2)/(1^4) = 0
E_z = (0)(3^2)/(1^3) - 4(0)(1^(1/2)) = 0
The magnitude of the electric field (E) is given by:
E = √(E_x^2 + E_y^2 + E_z^2)
E = √((-3)^2 + 0^2 + 0^2)
E = √9 = 3
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If a long wire is connected to the terminals of a Determine the current in the wire. 12−V battery, 9.6×10
19
electrons pass a cross Express your answer with the appropriate units. section of the wire each second. Part B Determine the resistance in the wire. Express your answer with the appropriate units. long wire is connected to the terminals of a Determine the power supplied by the battery. −V battery, 9.6×10
19
electrons pass a cross ction of the wire each second. Express your answer with the appropriate units. Part D Determine the power dissipated through the wire with the resistance found above. Express your answer with the appropriate units.
Given parameters:
Potential difference (V) = 12V
Electrons per second (n) = 9.6×10^19 electrons/s
Part A: Determine the current in the wire
The current I flowing in the wire is given by;
I = n × e × A Here, e = charge of one electron = 1.6 × 10⁻¹⁹ C = 1.6 × 10⁻¹⁹ As
and, A = area of cross-section of the wire = 1m²I = n × e × A= (9.6 × 10¹⁹) × (1.6 × 10⁻¹⁹) × (1)= 15.36 A
Thus, the current in the wire is 15.36 A.
Part B: Determine the resistance in the wire
The resistance R of the wire is given by;V = I × R ⇒ R = V / I = 12 V / 15.36 A= 0.78125 Ω
Thus, the resistance in the wire is 0.78125 Ω.
Part C: Determine the power supplied by the battery
The power supplied by the battery P is given by;P = V × I = 12 V × 15.36 A= 184.32 W
Thus, the power supplied by the battery is 184.32 W.
Part D: Determine the power dissipated through the wire with the resistance found above.
The power dissipated through the wire is given by;P = I² × R = (15.36 A)² × 0.78125 Ω= 183.085696 W
Thus, the power dissipated through the wire is 183.085696 W. Answer:Part A: 15.36 APart B: 0.78125 ΩPart C: 184.32 WPart D: 183.085696 W
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Person A lets go of an object at the same time Person B throws a similar object directly upward. The object thrown by Person B had enough speed that both objects will collide just as Person B's object hits its maximum height. Note that Person A is at yA,0=172 m; while, Person B is at YB,0=0 m. a. Find Person B's object's initial speed. b. Find the time of collision of both objects. c. What is the height where both objects collide with each other? d. After colliding, the objects were stuck together and moved directly downward. Their initial speed is u
0
=
2
1
gy
A,0
. Find the time after collision when both of the objects hit the ground
Person B's object's initial speed is 0 m/s. b. The time of collision of both objects is given as 6.20 seconds. c. The height where both objects collide is calculated as -120.34 meters.
To solve this problem, we can use the equations of motion and apply the principles of projectile motion.
a. To find Person B's object's initial speed, we need to determine the velocity at the maximum height, as this is the point where both objects collide. Since the object is thrown directly upward, the final velocity at the maximum height is 0. We can use the equation for vertical motion:
v = u + gt,
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (approximately -9.8 m/s^2), and t is the time taken to reach the maximum height. Substituting the known values, we have:
0 = u - 9.8t,
From the problem statement, we know that Person B's object collides with Person A's object just as it reaches its maximum height. Therefore, the time taken for Person B's object to reach the maximum height is the same as the time of collision.
b. To find the time of collision, we can equate the time taken by Person A's object to fall from the initial height to the time taken by Person B's object to reach its maximum height
t = sqrt(2h/g),
where h is the initial height. Substituting the known values, we have:
t = sqrt(2 * 172 / 9.8) = 6.20 s.
c. To find the height where both objects collide, we can use the equation for vertical motion:
h = u*t + (1/2)g*t^2,
where u is the initial velocity, g is the acceleration due to gravity, and t is the time of collision. Substituting the known values, we have:
h = 0 + (1/2)(-9.8)(6.20)^2 = -120.34 m.
The negative sign indicates that the height is below the initial reference point.
d. After colliding, the objects move directly downward with an initial speed of u0 = (1/2)g * yA,0. To find the time taken for both objects to hit the ground, we can use the equation for vertical motion:
h = u0*t + (1/2)g*t^2,
where h is the height (which is equal to -yA,0), u0 is the initial velocity, g is the acceleration due to gravity, and t is the time. Substituting the known values, we have:
-172 = (1/2)(-9.8)t + (1/2)(-9.8)t^2,
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = 0.5(-9.8), b = 0.5(-9.8), and c = -172.
Plugging in these values, we get:
t = (-0.5(-9.8) ± √((-0.5(-9.8))^2 - 4(0.5(-9.8))(-172))) / (2(0.5(-9.8))).
Simplifying further:
t = (4.9 ± √(24.01 + 332.8)) / (-9.8).
t = (4.9 ± √356.81) / (-9.8).
Taking the square root:
t ≈ (4.9 ± 18.88) / (-9.8).
This gives us two possible values for t:
t ≈ (4.9 + 18.88) / (-9.8) ≈ -2.15 seconds (not a valid solution since time cannot be negative).
t ≈ (4.9 - 18.88) / (-9.8) ≈ 1.47 seconds.
Therefore, after the collision, both objects will hit the ground approximately 1.47 seconds later.
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A Tomahawk cruise missile is launched from the barrel of a mobile missile launcher on a stationary submarine out in the ocean. The missile has a mass of 1315 kg and the submarine has a mass of 2×106 kg. The launch speed of the missile is 220 m/s. What is the speed of the submarine after the launch? 0.14 m/s 0 m/s 110 m/s 220 m/s
Therefore, the speed of the submarine after the launch is 0.14 m/s. A Tomahawk cruise missile has a mass of 1315 kg. The submarine on which the missile is launched has a mass of 2×[tex]10^6[/tex]kg.
The missile is launched from the barrel of a mobile missile launcher on a stationary submarine out in the ocean. The launch speed of the missile is 220 m/s. We have to find the speed of the submarine after the launch.
We have the following values:
Mass of missile, m1 = 1315 kg
Mass of submarine, m2 = 2 x [tex]10^6[/tex] kg
Launch speed of missile, u1 = 220 m/s
Velocity of submarine after launch, v2 = ?
As there is no external force on the submarine-missile system, therefore, the momentum of the system before the launch must be equal to the momentum of the system after the launch. So, we can write:
m1u1 + m2v2 = (m1 + m2)v1
Where v1 is the velocity of the system after the launch, which we have to calculate.
Substituting the given values, we get:
1315 x 220 + 2 x [tex]10^6[/tex] x v2 = (1315 + 2 x [tex]10^6[/tex]) x v1
Calculating the value of v1, we get:
v1 = (1315 x 220 + 2 x [tex]10^6[/tex]x v2) / (1315 + 2 x [tex]10^6[/tex])
After the missile is launched, its mass becomes zero. Therefore, the submarine will move with a certain velocity v2, which is its velocity after the launch. We can calculate it by substituting the above value of v1 and solving for v2:
v2 = 0.14 m/s
Therefore, the speed of the submarine after the launch is 0.14 m/s.
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A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s- until it reaches a speed of 25.0 m/s. Then the vehicle travels for 85.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the self-driving car in motion in s)? (b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.)
The self-driving car is in motion for 102.5 seconds.
The average velocity of the self-driving car for the described motion is approximately 23.80 m/s.
(a) To find the total time the self-driving car is in motion, we need to sum up the time it takes to accelerate, the time it travels at a constant speed, and the time it takes to decelerate.
Given:
Acceleration (a) = 2.00 m/s²
Final speed (vf) = 25.0 m/s
Time to decelerate (t_deceleration) = 5.00 s
Time at constant speed (t_constant) = 85.0 s
Time to accelerate:
We can use the formula of motion to find the time it takes to accelerate from rest to a final speed:
vf = vi + at
Since the initial velocity (vi) is 0 m/s:
25.0 m/s = 0 m/s + (2.00 m/s²)t
Solving for t:
t = 25.0 m/s / 2.00 m/s²
t ≈ 12.5 s
Total time in motion:
The total time in motion is the sum of the time to accelerate, the time at constant speed, and the time to decelerate:
Total time = t_acceleration + t_constant + t_deceleration
Total time = 12.5 s + 85.0 s + 5.00 s
Total time = 102.5 s
Therefore, the self-driving car is in motion for 102.5 seconds.
(b) To find the average velocity of the self-driving car, we need to calculate the total displacement and divide it by the total time.
Total displacement consists of two parts: the distance covered during acceleration and deceleration and the distance covered at constant speed.
Displacement during acceleration and deceleration:
We can use the formula of motion to find the displacement during acceleration:
vf = vi + at
Since the initial velocity (vi) is 0 m/s:
25.0 m/s = 0 m/s + (2.00 m/s²)t
Solving for t:
t = 25.0 m/s / 2.00 m/s²
t ≈ 12.5 s
Using the formula for displacement during uniform acceleration:
d = vi * t + (1/2) * a * t²
Where:
vi is the initial velocity (0 m/s)
t is the time (12.5 s)
a is the acceleration (2.00 m/s²)
d_acceleration = 0 * 12.5 + (1/2) * 2.00 * (12.5)²
Simplifying the equation:
d_acceleration = (1/2) * 2.00 * 156.25
d_acceleration = 156.25 m
The displacement during deceleration is the same as during acceleration:
d_deceleration = 156.25 m
Displacement at constant speed:
The displacement at constant speed can be calculated using the formula:
d_constant = v * t
Where:
v is the constant speed (25.0 m/s)
t is the time at constant speed (85.0 s)
d_constant = 25.0 m/s * 85.0 s
d_constant = 2125 m
Total displacement:
The total displacement is the sum of the displacements during acceleration, deceleration, and at constant speed:
Total displacement = d_acceleration + d_constant + d_deceleration
Total displacement = 156.25 m + 2125 m + 156.25 m
Total displacement = 2437.5 m
Finally, we can calculate the average velocity:
Average velocity = Total displacement / Total time
Average velocity = 2437.5 m / 102.5 s
Calculating the average velocity gives us:
Average velocity ≈ 23.80 m/s
Therefore, the average velocity of the self-driving car for the described motion is approximately 23.80 m/s.
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