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andrey2020 [161]
2 years ago
14

A 20 N force pulls to the right and friction pulls 5 N. If the mass is 5 kg, find acceleration.​

Physics
1 answer:
Slav-nsk [51]2 years ago
7 0

Answer:

Explanation:

Force - frictional Force = m * a

20N - 5N = 5 * a

15N = 5*a

15N / 5 = a

a = 3 m/s^2

You should note a couple of things.

1. The frictional Force always opposes the direction of motion. In this case it is minus and the 20 N is plus. It makes the acceleration less. If the friction force was not there then 20 N = m * a

20N = 5 kg * a

a = 20/5

a = 4 m/s^2.

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The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

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Mathematically,

\frac{T^2}{a^3}=const.

Where

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3 years ago
You observe lighting striking and then hear to sound 8 seconds later. The speed of sound in air is 340 m/s. How
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2 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
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Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

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Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

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but,

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where,

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Comparing both equations:

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v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

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m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

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<u>v = 2149.24 m/s</u>

<u></u>

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