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AleksandrR [38]
3 years ago
15

Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh

at distance will it cover during that time
Physics
1 answer:
Nutka1998 [239]3 years ago
4 0

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s

Let d is the distance cover during that time. So,

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

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A projectile is shot on level ground with a
erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m    agreeing with answer found above

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I’m so confused. please help. i don’t know what i’m suppose to do
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Explanation:

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A young girl slides down a rope. As she slides
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Answer:

The girl will move with constant velocity

Explanation:

If after a certain time t_0 the velocity of the girl is v_0 =gt_0 and the upward force on the girl due to rope is mg ,where g is gravitational acceleration. Then the girl will move down with the constant velocity v_0 .

The girl will move with constant velocity,as explained above.

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Ground-based radio telescopes can collect data from distant objects in space
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Read 2 more answers
A girl throws a pebble into a deep well at 4.0 m/s (downward). It hits the water in 2.0 s. How far below the ground is the water
alexandr402 [8]

Answer:

  • 27.6 m
  • 13.8 m/s

Explanation:

(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...

  4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity

__

(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...

  d = vt

  d = (13.8 m/s)(2 s) = 27.6 m

The water is about 27.6 m below ground.

_____

* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:

  vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2

__

If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

  \displaystyle d=\int_0^t{(v_0+at)}\,dt=v_0t+\dfrac{1}{2}at^2=t\left(v_0+a\dfrac{t}{2}\right)\\\\v_{avg}=\dfrac{d}{t}=v_0+a\dfrac{t}{2}\qquad\text{the formula we started with}

8 0
3 years ago
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