All categories

4 years ago

A box falls out of a stationary helicopter hovering 135 m above the ground. How long will it take to hit the ground?

1 answer:

4 0

We have the equation of motion $s = ut + \frac{1}{2} at^2$ , where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here displacement = 135 m, Initial velocity = 0 m/s, acceleration = 9.81 $m/s^2$

Substituting

$135 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 =135\\ \\ t =5.25 seconds$

A box falls out of a stationary helicopter hovering 135 m above the ground will take 5.25 seconds to reach the ground.

You might be interested in

A girl starts from rest and reaches a walking speed of 1.4 m/s in 3.0 s. She walks at this speed for 6.0 s. The girl then slows

Alona [7]

Answer:

1) The girl's acceleration at time 't' is, $a=0.47$ m/s²

2) The girl's acceleration at time 't₁' is, $a_{1}=0$ m/s²

3) The girl's acceleration at time 't₂' is, $a_{2}=-0.14$ m/s²

Explanation:

Given data,

The initial walking speed of the girl, u = 0

The speed of the girl at the time 't' 3 s is, v₁ = 1.4 m/s

The time period the girl walked at the speed 1.4 m/s is, t₁ = 6 s

The girl slows down and comes to a stop during a period, t₂ = 10 s

1) The girl's acceleration at time 't'

$a=\frac{v-u}{t}$

$a=\frac{1.4-0}{3}$

$a=0.47$ m/s²

2) The girl's acceleration at time 't₁'

$a_{1} =\frac{v_{1} -v}{t_{1}}$

$a_{1}=\frac{1.4-1.4}{6}$

$a_{1}=0$ m/s²

3) The girl's acceleration at time 't₂'

$a_{2} =\frac{v_{2} -v_{1}}{t_{2}}$

$a_{2}=\frac{0-1.4}{10}$

$a_{2}=-0.14$ m/s²

4 0

3 years ago

John is hiking and notices a small stream of water flowing down the side of the mountain. What part of the water cycle is John o

Alexus [3.1K]

Answer: Runofff

Explanation: because it said that it is going down the side of the mountain

3 0

3 years ago

A truck driver is going down a one way street the wrong way, and passes at least ten cops. Why is he not caught?

garik1379 [7]

I think because there is only one way to go

6 0

3 years ago

A tire is filled with air at 10 ∘C to a gauge pressure of 250 kPa. Part A If the tire reaches a temperature of 45 ∘C, what fract

Alex_Xolod [135]

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = $10^{\circ}$ = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = $45^{\circ}$ = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

m = no. of moles of gas

R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

$\frac{m}{m'} = \frac{T'}{T}$ (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

$y = \frac{m - m'}{m} = 1 - \frac{m'}{m}$

From eqn (1):

$y = 1 - \frac{T}{T'}$

$y = 1 - \frac{283}{318} = 0.11$

6 0

4 years ago

I need a bit of help on this one, can anyone answer?

emmainna [20.7K]

Potential energy = m · g · h

-- When you held the ball at 2.0 meters above the floor, it had

(0.5 kg) · (9.8 m/s²) · (2.0 m) = 9.8 Joules of potential energy.

-- After it bounced and went back up as high as it could, it was only 1.8 meters above the floor. Its potential energy was

(0.5 kg) · (9.8 m/s²) · (1.8 m) = 8.82 Joules

-- Between the drop and the top of the bounce, it lost

(9.8 - 8.82) = <em>0.98 Joule</em> .

-- The energy was lost when the ball hit the floor. During the hit, 0.98 joule of kinetic energy turned to <em>thermal energy</em>, which slightly heated the ball and the floor.

5 0

3 years ago