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zhuklara [117]
3 years ago
6

If you swing an object slowly, it may rotate at less than one revolution per second. What would be the revolutions per second fo

r an object that makes one revolution in five seconds

Physics
2 answers:
igomit [66]3 years ago
4 0

Answer:

0.2 revolution per second

Explanation:

If the object makes 1 revolution in every 5 seconds, then for each second, it would make an amount of a revolution divided by 5, or 0.2 revolution. This is less than one revolution per second because you are spinning slowly and it takes more than 1 second to complete 1 revolution

Ierofanga [76]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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What two parts make up velocity?
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Speed and direction are the two scientific componets that make up velocity.

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3 years ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

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3 years ago
A + 32.2 μC charge feels a 0.544 N force from a + 12.3p μC charge. How far apart are they? (u stands for micro.) [?] m
love history [14]

Answer:

r = 2.55 m

Explanation:

Given that,

First charge, q₁ = 32.2 μC

Second charge, q₂ = + 12.3 μC

The force between charges, F = 0.544 N

We need to find the distance between charges. The force between two charges is given by the formula as follows :

F=k\dfrac{q_1q_2}{r^2}\\\\r=\sqrt{\dfrac{kq_1q_2}{F}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 32.2\times 10^{-6}\times 12.3\times 10^{-6}}{0.544}} \\\\r=2.55\ m

So, the charges are 2.55 m apart.

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