It’s 4. Definitely 4. 99% sure it’s 4.
Answer:
<u>note:
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<u>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
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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
Answer:
106.7 N
Explanation:
We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:
where
F is the average force
is the duration of the collision
m is the mass of the ball
v is the final velocity
u is the initial velocity
In this problem:
m = 0.200 kg
u = 20.0 m/s
v = -12.0 m/s
Solving for F,
And since we are interested in the magnitude only,
F = 106.7 N
