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alukav5142 [94]
3 years ago
13

If a ball is 10m high with what velocity will it fall?

Physics
1 answer:
Semmy [17]3 years ago
6 0

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

  V² = U² + 2gH

where V is the final velocity

            U is the initial velocity

             g is the acceleration due to gravity 9.8m/s²

             H is the height of fall

The initial velocity here is zero and

      V² = 2 x 9.8 x 10 = 196

       V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

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A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this
olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is 3442.04 m

Explanation:

Given that the rocket moves with an acceleration a=30m/s^2

time t=5 s

Since the rocket starts from rest initial velocity  u=0 s

The distance it travelled within this time is given by  s=ut+ \frac{1}{2} at^2                                                                                                  =0 \times 5+ \frac{1}{2} (30\times25)=375 m

Velocity at this point is given by v=u+at

v=0+30\times5=150m/s

Given that at this height it runs out of fuel but travels further. Here final velocity v=0(maximum height), initial velocityu=150 m/s  and time to zero velocity t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.

Thus it travels 15.3 seconds more after fuel running out. The distance covered during this period is given

s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m

7 0
2 years ago
Derive the formula 2as=v² -u² where the formula have have usual meanings
attashe74 [19]

Answer:The main formula is v² = u² + 2as

Explanation:

S=½(u +v)t

t = v+u/a

S=½(v-u)(v+u/a)

S=½v²+uv-uv-u²/a

2as=v²-u²

5 0
3 years ago
The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated
Sedaia [141]

Answer:

60

Explanation:

According to the given question, the computation of minimum coating thickness is shown below:-

The condition for constructive interference is

2t_{min} = (m + \frac{1}{2} )\times \frac{\lambda}{^nmateral}

= (0 + \frac{1}{2} )\times \frac{\lambda}{^nmateral}

t = \frac{\lambda}{4n}

Now we will put the values to the above formula to reach the answer

= \frac{480nm}{4\times 2.0}

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Therefore we simply applied the above formula to determine the minimum coating thickness

6 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

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In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

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r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
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What is the relationship between thickness of lens and focal length?​
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Thick lens will have shorter and consequently thin lens will have greater focal length. Because, For a thick lens, the optical path length of the light is more, than for a thin lens, thus, the bending of light will be more in case of a thicker lens. Consequently, it has a shorter focal length.

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