Solution
distance travelled by Chris
\Delta t=\frac{1}{3600}hr.
X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}
=\frac{579.5}{3600}=0.161miles
Kelly,
\Delta t=\frac{1}{3600}hr.
X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}
=\frac{657.5}{3600}
\Delta X=X_{k}-X_{C}=0.021miles
The net force of the object is equal to the force applied minus the force of friction.
Fnet = ma = F - Ff
12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107.
The spring constant is 147 N/m
Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm
We need to find the spring constant
A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring
We know that F = kx
300(9.8)= k (0.02)
k = 147.15 N/m
Rounding off to the nearest is 147N/m
The spring constant is 147N/m
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Newton's third law of motion
Explanation:
Newton's third law of motion states that:
<em>"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force (reaction force) on object A"</em>
It is important to note that this law is always valid, even when it seems it is not.
Consider for example the gravitational force that the Earth exerts on your body (= your weight). We can say that this is the action force. It may seems that there is no reaction force in this case. However, this is not true: in fact, your body also exerts an equal and opposite force on the Earth, and this is the reaction force. The reason that explains why we don't notice any effect on Earth due to this force is that the mass of the Earth is much larger than your mass, therefore the acceleration produced on the Earth because of the force you apply is negligible.
It is also important to note that the action-reaction pair of forces always act on two different objects, so they never appear in the same free-body diagram.
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In a constant acceleration of 3m per second, after 10 seconds,
3 x 10 = 30
B. 30m/s is your answer
hope this helps :D
