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Alika [10]
3 years ago
9

A 55 kg skater at rest on a frictionless rink throws a 3 kg ball, giving the ball a velocity of 8 m/s. What is the velocity of t

he skater immediately after?
Physics
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

The velocity will be v1 = 0.58[m/s]

Explanation:

This problem can be solved by the law of conservation of the moment, which explains that the moment of a system remains constant because there are no external forces acting on it.

We have the following initial data:

m1 = mass of the skater = 55 [kg]

m2 = mass of the ball = 3 [kg]

v2 = velocity of the ball = 8 [m/s]

Therefore:

m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}+m_{2}*v_{2}\\(50*0)+(3*0)=(50*v_{1})+(3*8)\\50+3-24=50*v_{1}\\v_{1}= 0.58[m/s]

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5 0
3 years ago
How can electric energy be converted to heat energy?
Goshia [24]
To do that, you must pass electric current through a substance
that electrons have to spend energy to pass through. 
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We'll say that the substance has "resistance", which we can measure.
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5 0
3 years ago
A technician services the carburetor, and then, performs a complete governor system adjustment. The governor system on the engin
madreJ [45]

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5 0
3 years ago
At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

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= g1 ×1/2

= g1/2

6 0
3 years ago
What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?
Andru [333]
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26} protons

Then, the total charge is the number of protons times the charge of a  single proton:
Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C
8 0
3 years ago
Read 2 more answers
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