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melamori03 [73]
3 years ago
13

What are the two factors that affect the frictional force between two surfaces?

Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer:

The amount of force and the angle between them.

Explanation:

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A college student is working on her physics homework in her dorm room. Her room contains a total of 6.0 x 10^26 gas molecules. A
IceJOKER [234]

Answer:

Temperature, T = 3.62 kelvin

Explanation:

It is given that,

Total number of gas molecules, N=6\times 10^{26}

Her body is converting chemical energy into thermal energy at a rate of 125 W, P = 125 W

Time taken, t = 6 min = 360 s

Energy of a gas molecules is given by :

\Delta E =\dfrac{3}{2}NkT

T=\dfrac{2E}{3Nk}, k is Boltzmann constant

T=\dfrac{2\times P\times t}{3Nk}

T=\dfrac{2\times 125\times 360}{3\times 6\times 10^{26}\times 1.38\times 10^{-23}}

T = 3.62 K

So, the temperature increases by 3.62 kelvin. Hence, this is the required solution.

4 0
3 years ago
The atmosphere protects us from _____ and _____.
sveticcg [70]

Answer: B. meteorites, UV rays

The earth's atmosphere plays an important role in shielding the space rocks and meteorites enters inside the earth due to collision with the celestial body. Due to high heat and pressure in the mesospheric layer of the atmosphere these meteorites burned up before reaching the biosphere or hydrosphere system of the earth. The UV rays are protected from entering into the earth atmosphere by the ozone layer present in the stratosphere of the atmosphere. Both meteorites and UV rays can negatively effect the human population. Therefore, the atmosphere is an agent which provides protection against them.

7 0
3 years ago
Read 2 more answers
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m
Nataliya [291]
Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
8 0
3 years ago
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m
umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s

v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads

Now, we just use the formula:

w_f^2-w_o^2=2\alpha*x

\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2

6 0
3 years ago
Read 2 more answers
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