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xeze [42]
2 years ago
6

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction betwe

en the puck and ice is 0.11, how much time does it take the puck to come to rest?
Physics
1 answer:
Salsk061 [2.6K]2 years ago
5 0
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

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What is the difference between a bound orbit and an unbound orbit around the sun?
Anit [1.1K]

Answer:

Explanation:

The difference between a bound orbit and an unbound orbit around the sun is that:

An object on a bound orbit pursues the same way around the Sun again and again, while an object on an unbound orbit moves toward the Sun only a single time and afterward stays away forever &  never returns.

8 0
3 years ago
Select all of the answers that apply. Jenna dives 20meters into the ocean. How much pressure does she feel?
Triss [41]
Use the formula dgh + p atm

d is density g is gravitational field strength h heigh or depth and p atm is atmospheric pressure

1000 x 10 x 20 + 100000 = 300000

almost 300,000 pascal
4 0
4 years ago
This is a machine that converts electrical energy into mechanical energy.
Yuri [45]
<h2>Answer:</h2>

Motor

<h2>Explanation:</h2>

A motor is a machine that converts electrical energy into mechanical energy. In motors, electric energy is converted into mechanic energy when a magnetic torque acts on a conductor that carries a current. There are different types of motors like DC and AC motors. The moving part of a motor is called the rotor while the stationary part is called stator

8 0
3 years ago
If a car's mass is 439 kg, what is its weight on earth?
trasher [3.6K]
The correct answer is B.
8 0
3 years ago
A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn
kow [346]

Answer:

The tension in the cord is T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} } (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

Mg=(M+\frac{MR^{2} }{2b^{2} }  )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2}  }

Replacing in eq. 1

T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

3 0
3 years ago
Read 2 more answers
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