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2 years ago

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A 1-kg mass is dropped from a third floor window. The acceleration of the mass is found to be 8 m/s2. What is the average force

of air resistance? A) 1.8 N B) 7 N C) 8 N D) 9.8 N

1 answer:

Paha777 [63]2 years ago

7 0

Summary:
m=1kg
a=8 m/s^2
g= 9,8 m/s^2
F(ar)=?

I hope to help you

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A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum".

Alex787 [66]

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

Part d)

$W = 1012.7 J$

Part e)

$v = 2.1 m/s$

Part f)

$E = 1037.2 J$

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

$\theta = 38^o$

so the maximum height reached by the pendulum is given as

$h = L(1 - cos\theta)$

so we will have

$h = L(1 - cos38)$

$h = 1.25(1 - cos38)$

$h = 0.265 m$

now gravitational potential energy of the pendulum is given as

$U = mgh$

$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

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

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.265)}$

$v = 2.28 m/s$

Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

$0.065 v = (5 + 0.065)2.28$

$v = 177.66 m/s$

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

$W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2$

$W = 1012.7 J$

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

$0 = mv_1 + Mv_2$

$0 = 0.065(177.6) + 5.50 v$

$v = 2.1 m/s$

Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$

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2 years ago

Why is it unwise to stir a pot of soup with a metal spoon?

emmasim [6.3K]

Answer:

Explanation:

As we stir the pot of soup by a metal spoon which is a conductor, it conducts the heat and starts heating after some time the temperature of the spoon rises and we are not able to hold the spoon and we get burns.

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2 years ago

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Mrs. Brown's class is studying magnets and electricity. At the end of the unit Claire states that magnets and electricity are bo

Gwar [14]

Answer:

Magnets can create electricity and electricity can create a magnetic force.

Both electric charges and magnets do not have to touch an object in order to exert a force on it.

Electromagnets use electricity to create a magnetic force.

Explanation:

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2 years ago

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A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

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

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

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2 years ago

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

$W = q \Delta V$

where

$q=3.0 \mu C = 3.0 \cdot 10^{-6} C$ is the magnitude of the charge

$\Delta V = 6.0 kV = 6000 V$ is the potential difference between point P and infinity

Substituting into the equation, we find

$W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J$

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2 years ago