The magnitude of the force of friction is 40 N
Explanation:
To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:
- The constant pushing force forward, of magnitude F = 40 N
- The frictional force, acting backward, 
Since the two forces are in opposite direction, the equation of motion is
where
m is the mass of the student+scooter
a is the acceleration
However, here the scooter is moving at constant speed: this means that its acceleration is zero, so
a = 0
And therefore,
which means that the magnitude of the force of friction is also equal to 40 N.
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A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
The speed at which the objects were attracted to each other determined by the gravitational pull.
<h3>What is gravity?</h3>
The force of attraction felt by a person at the center of a planet or Earth is called as the gravity or gravitational pull.
The gravitational force, Fg is the attractive force exerted by the Earth on the object that is equal to the mass of the object times the gravitational acceleration.
When the two objects are attracted, the speed with which they are moving toward each other depends on the strength of the pull force. This pull force is the gravitational pull.
Thus, the speed at which the objects were attracted to each other determined by the gravitational pull.
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Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
The answer is the letter "C" ( I have honors science I am good at this type of stuff )
Hope I helped :) ( ask me for help when u need it :)
