For an object to be in projectile motion, what force must be acting on it

Which equations can be used to solve for acceleration? Select four options.

tatiyna

Explanation:

Acceleration of an object is calculated by finding the change in its velocity divided by time taken.

If $v_i$ is initial velocity, $v_f$ is final velocity and t is time taken. Then the acceleration of the object is given by :

$a=\dfrac{v_f-v_i}{t}\\\\at=v_f-v_i\\\\v_f=v_i+at$ .....(1)

So, the above equation is used to find acceleration. It is called the first equation of motion. After rearranging equation (1), the correct options are :

$t=\dfrac{\Delta v}{a}$

$v_i=v_f-at$

$v_f=v_i+at$

$a=\dfrac{\Delta v}{t}$

7 0

3 years ago

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1. The uniform seesaw is balanced at its center of mass. The smaller boy on the right has a mass of m = 40.0 kg. What is the mas

tankabanditka [31]

Answer:

Explanation:

Find the complete question attached

Using the principle of moment

Clockwise moment = Anticlockwise moment

AntiClockwise moment = M × 2.0

ACW moment = 2M

Clockwise moment = 40×4

Clockwise moment = 160kgcm

Equate both expression and calculate M

2M = 160

M = 160/2

M = 80kg

Hence the mass of his friend is 80kg

4 0

4 years ago

IF YOU GOOD AT SCIENCE PLEASE ANSWER THIS ASAP

xz_007 [3.2K]

Answer:

I would believe that it would be the last option

Explanation:

Physical science is a type of science that mainly focuses on natural objects that are not alive, such as minerals and rocks.

4 0

3 years ago

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((((Psychology))))According to Rubin, the desire to share your innermost thoughts and feelings is:

victus00 [196]

According to Rubin, the desire to share your innermost thoughts and feelings is

c. intimacy

Hope this helps.

5 0

3 years ago

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

Nina [5.8K]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

Now the resultant of the two velocities perpendicular to each other:

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

3 0

4 years ago