Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 3 s. What is your acceleration?

Please help asap! Thank you.

olganol [36]

Answer:

direct current

Explanation:

it has a direct path to go down to reach the specific point

4 0

3 years ago

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2. Tony sets up an experiment inside his classroom. He spaces 5 plants at equal

brilliants [131]

Answer:

a

Explanation:

4 0

3 years ago

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9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

3 years ago

Amir and Gracie left some evidence of breakfast behind. Luckily, some of it can be changed back. What does Mosa write down as cl

marta [7]

Answer:

You could say that it was not a chemicly change because that is when you say this is just a example that if you cook eggs you cant un-cook them

Explanation:

3 0

2 years ago

shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r

Ainat [17]

Answer:

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0$

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(3R-R)}{(R^2+4R^2)^{3/2}}=0$

$\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0$

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

8 0

3 years ago

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