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Mariulka [41]
3 years ago
13

Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

ond. Find your deceleration and estimate the force (in newtons) that the wall exerted on you during the stopping?
Physics
1 answer:
STatiana [176]3 years ago
3 0

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration=\frac{v-u}{t}

Using the formula

Acceleration=a=\frac{0-4.5}{0.5}

deceleration=a=-9m/s^2

We know that

Force =ma

Using the formula and suppose mass  of my body=m=40 kg

The force exerted by wall on you

Force=40\times (9)=360N

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Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

                   \vec{A}=x_A\hat{i}

Vector B , which has a magnitude of 5.5 m

                   \vec{B}=x_B\hat{i}+y_B\hat{j}

                   \sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A                    \vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}

Comparing we will get

                  x_A=-x_B\\\\y_B=6x_A

Substituting in x_{B}^{2}+y_{B}^{2}=30.25

                  \left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904

So we have

    \vec{A}=0.904\hat{i}

Magnitude of vector A = 0.904

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