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3 years ago

A cylindrical object can roll down an incline, as shown in Figure 1. The incline is slightly less

Physics

1 answer:

Olin [163]3 years ago

4 0

Answer:

a = 2d / t²

a = 2gh / (3d)

Explanation:

One method is to use the equation:

Δx = v₀ t + ½ at²

d = (0) t + ½ at²

d = ½ at²

a = 2d / t²

By measuring the length of the incline d, and the time it takes to reach the bottom t, the students can calculate the acceleration, using only the meter stick and the stopwatch.

Another method is to use conservation of energy to find the final velocity.

Initial potential energy = final rotational energy + kinetic energy

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For a solid cylinder, I = ½ mr². For rolling without slipping, ω = v/r.

mgh = ½ (½ mr²) (v/r)² + ½ mv²

mgh = ¼ mv² + ½ mv²

mgh = ¾ mv²

4gh/3 = v²

Using constant acceleration equation:

v² = v₀² + 2aΔx

4gh/3 = 0² + 2ad

a = 2gh / (3d)

Using this equation, the students can measure the height of the incline h, and the length of the incline d, to calculate the acceleration. The only equipment needed is the meter stick.

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4. A train starts its journey and accelerates at 5 ms 2. How long does it take for it to reach a velocity of 100 ms??

Agata [3.3K]

Answer:

20 seconds

Explanation:

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4 years ago

Read 2 more answers

at the submicroscopic and microscopic levels temperature is proportional to the average ______ kinetic energy of the particles o

In-s [12.5K]

Answer:

Explanation:

At the submicroscopic and microscopic levels temperature is proportional to the average degree of kinetic energy of the particles of a substance

8 0

3 years ago

Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change

Ket [755]

Answer:

a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

Explanation:

Let consider the bottom of the first floor in a building as the zero reference ( $z = 0\,m$ ). The change in potential energy experimented by a particle ( $\Delta U_{g}$ ), measured in joules, is:

$\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$z_{o}$ , $z_{f}$ - Initial and final height with respect to zero reference, measured in meters.

Please notice that $m\cdot g$ is the weight of the particle, measured in newtons.

a) If we know that $m = 3\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{o} = 0\,m$ and $z_{f} = 4.1\,m$ , then the change in potential energy is:

$\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)$

$\Delta U_{g} = 120.626\,J$

The change in potential energy of a 3.0 kilograms backpack carried up the stairs.

b) If we know that $m\cdot g = 650\,N$ , $z_{o} = 4.1\,m$ and $z_{f} = 0\,m$ , then the change in potential energy is:

$\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)$

$\Delta U_{g} = -2665\,J$

The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.

5 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

For general projectile motion, when the projectile is at the highest point of its trajectory, its acceleration is zero? A)The ho

Daniel [21]

Answer:

B) Its velocity is perpendicular to the acceleration.

Explanation:

For general projectile motion, the horizontal acceleration is 0 and the vertical acceleration is -g. This is true for all points on the trajectory.

At the highest point, the vertical velocity is 0. So you have only a horizontal velocity as well as a vertical acceleration. So the two are perpendicular.

4 0

4 years ago