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Zarrin [17]
2 years ago
5

A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is

0.7 meters above the floor, its potential energy exactly equals its kinetic energy. How fast is the flashlight moving at this point, in m/s 3.7
Physics
1 answer:
alexgriva [62]2 years ago
8 0

Answer:

The speed of the flashlight at that point is 3.7 m/s

Explanation:

When an object of mass M is at a height H above the ground, the potential energy of the object is:

U = M*H*g

Where g is the gravitational acceleration, g = 9.8 m/s^2

And for an object with velocity v, the kinetic energy is:

K = (M/2)*v^2

We know that when the flashlight of mass  1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:

M = 1kg

H = 0.7m

g = 9.8 m/s^2

Replacing these in the equations, we get:

U = K

(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2

As the mass factor appears in both sides, we can remove it:

(0.7 m)*(9.8 m/s^2) = (v^2)/2

Now we can multiply both sides by 2:

2*(0.7 m)*(9.8 m/s^2) = v^2

Now let's apply the square root to both sides:

√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s

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Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

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3 years ago
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8 0
1 year ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
3 years ago
The time period T of a simple pendulum is given by the relation
Vanyuwa [196]

Answer:

T^2 \propto L

Explanation:

The period of a simple pendulum is given by:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

From this equation we can write

T\propto \sqrt{L}\\T\propto \frac{1}{\sqrt{g}}

Taking the square of this equation, we get:

T^2 = (2\pi)^2 \frac{L}{g}

So we see that T^2 is proportional to L and inversely proportional to g. So, we can write:

T^2 \propto L\\T^2 \propto \frac{1}{g}

So the only correct option is

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Gennadij [26K]

Let the height where we are trapped is H

now to find the time to reach the key at the bottom is given as

y = v_i t + \frac{1}{2}at^2

now we have

H = \frac{1}{2}gt^2

t = \sqrt{\frac{2H}{g}}

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now by solving above equation we have

H = 48 m

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so our position must be

N = \frac{48}{3} = 16 th \:floor

3 0
3 years ago
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