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Mrrafil [7]
3 years ago
13

A crate falls from an airplane flying horizontally at an altitude of 2,000 m.

Physics
1 answer:
Hatshy [7]3 years ago
7 0

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

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describe the time of day that an early explorer might have planned to enter a harbor and when he might have planned to leave
inessss [21]
He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.
4 0
3 years ago
Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that
Tresset [83]

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

8 0
3 years ago
Read 2 more answers
A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
ValentinkaMS [17]

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

4 0
3 years ago
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0
3 years ago
An object moving north with an initial velocity of 14 m/s accelerates 5 m/s^2 for 20 seconds.what is the final velocity of the o
bekas [8.4K]

Answer:

Option C = 114 m/s

Explanation:

a=v-u/t

By substituting,

5=v-14/20

100=v-14

Thus, v=100+14

v= 114 m/s

Hope it helps :)

5 0
3 years ago
Read 2 more answers
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