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2 years ago

5

Aperson walk along actual path of radias 5m/s in 10 second what is the acceleration

1 answer:

AlekseyPX2 years ago

4 0

Answer:50m/s

Explanation:

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A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east

Marrrta [24]

Answer:

ax = 6.43m/s²

Explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0

3 years ago

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

3 years ago

Zak, helping his mother rearrange the furniture in their living room, moves a 60 kg sofa 5.9 m with a constant force of 40 N. Wh

Len [333]

Answer:236 J

Explanation:

Given

Mass of sofa $m=60 kg$

Displacement $s=5.9 m$

Force $F=40 N$

neglecting Friction

Work done is given by

$W=F\cdot s$

$W=F\cdot s\cos 0$

$W=40\cdot 5.9$

$W=236 J$

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3 years ago

Read 2 more answers

The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the

Margaret [11]

Answer:

$v=2019.09\ m.s^{-1}$

Explanation:

Given:

time taken by the sun to complete one revolution, $t=601.2\ hr$

radial distance of the sunspot, $r=6.955\times 10^8\ m.s^{-1}$

<u>Therefore, angular speed of rotation of sun:</u>

$\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}$

<u>Now the tangential velocity of the sunspot can be given by:</u>

$v=r.\omega$

$v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}$

$v=2019.09\ m.s^{-1}$

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3 years ago

The picture below shows the setup for an experiment involving a 1000 mL beaker, sitting on a burner, filled with 500 mL of water

Vlad1618 [11]

Answer:

bye

Explanation:

bye

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3 years ago