Answer:
distance= velocity ×time
distance= 62×10
distance=620m
hope it helps you mate please mark me as brainliast
Answer:
Explanation:
Recall the formula for acceleration:
, where 
is final velocity, 
is initial velocity, and 
is elapsed time (change in velocity over this amount of time).
Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.
We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).
We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.
Substituting values in our formula, we have:
Alternative:
Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!
Answer:
r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Answer:
According to <em>Newton's first law of motion:</em>
<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>
<u>It stays in motion with the same speed and goes in the same direction.</u>
<u></u>
<em>Hope this helped </em>
<em>:)</em>
The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object is 16.4 N.
<h3>What is frictional force?</h3>
Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.
The normal force acting on the object of mass 4.2 Kg is N = mg
N = 4.2 Kg × 9.8 m/s² = 41.16 N
Frictional force = ц N
= 0.40 × 41.16 N
= 16.4 N.
Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N
To find more on frictional force, refer here:
brainly.com/question/1714663
#SPJ1
Your question is incomplete. But your complete question probably was:
The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?
