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4 years ago

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

Physics

1 answer:

Hatshy [7]4 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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Which elements are metals?Check all that apply.

ELEN [110]

HG. Li. Re dnsdfjnjdfsnijfdsnjfnsdjifnijdsnfijdnsijfijs

3 0

3 years ago

How to solve part (b)?

maria [59]

Answer:

1.11m

Explanation:

From the diagram we are given the following forces;

F1 = 24.3N

F3 = 30N

Since the sum of upward forces is equal to that of downward force, then;

F2 = F1 + F3

F2 = 24.3N + 30N

F2 = 54.3N

Required

Distance between B and C

First we need to get Length of AC

Take moment about A

Anticlockwise moment = F3 cos20 * AC

Anticlockwise moment = 30ACcos 20

Clockwise moment = 1.2 * F2

Clockwise moment = 1.2(54.3) = 65.16Nm

Applying the principle of moment;

Sum of ACW moment = Sum of CW moments

30ACcos 20 = 65.16

AC = 65.16/30cos20

AC = 65.16/28.19

AC = 2.31m

Get the distance BC

AC = AB + BC

BC = AC-AB

BC = 2.31 - 1.2

BC = 1.11m

Hence the separation between B and C is 1.11m

Note that the force F1 got in (a) was the value used in the calculation.

8 0

3 years ago

I'll mark brainless pictures down below

julsineya [31]

The net force is (40 Newtons) (down the road).

But 40 Newtons is not going to move a piano very enthusiastically.

8 0

3 years ago

Kraig pulls a box to the right at an angle of 40 degrees to the horizontal with a force of 30 Newtons. If Kraig pulls the box a

eimsori [14]

Answer:

459.6J

Explanation:

Given parameters:

Angle of pull = 40°

Force applied = 30N

Distance moved = 20m

Unknown:

Work done by Kraig = ?

Solution:

To solve this problem;

Work done = F x dcosФ

d is the distance

F is the force

Ф is the angle given

Work done = 30 x 20cos40° = 459.6J

3 0

3 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago