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3 years ago

8

If earth had no atmosphere , would a falling object ever reach terminal velocity?

1 answer:

Bas_tet [7]3 years ago

6 0

The right answer is no.

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By the nucleus, protons and neutrons of an atom does it contain more mass

Temka [501]

Answer:

Virtually all the mass of an atom resides in its nucleus, according to Chemistry

Explanation:

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3 years ago

A force of 25 N acts on an object at an angle of 30 degrees above the horizontal. How much work does the force do if the object

8_murik_8 [283]

We can find the x-component of the force vector, so Fx = 25N(cos30). Then, we can sub this into the work equation W = Fs or W = Fd.
Therefore, W = (25N(cos30))(3m) = 64.95 J.
With sig figs, it is 60 J.

3 0

3 years ago

To win the game, a place kicker must kick a

masya89 [10]

Answer:

0.57 m

Explanation:

First of all, we need to calculate the time it takes for the ball to cover the horizontal distance between the starting position and the crossbar. This can be done by analzying the horizontal motion only. In fact, the horizontal velocity is constant and it is

$v_x = u cos \theta = (15)(cos 51.7^{\circ})=9.30 m/s$

And the distance to cover is

d = 19 m

So the time taken is

$t=\frac{d}{v_x}=\frac{19}{9.30}=2.04 s$

Now we want to find how high the ball is at that time. The initial vertical velocity is

$u_y = u sin \theta = (15)(sin 51.7^{\circ})=11.77 m/s$

So the vertical position of the ball at time t is

$y(t) = u_y t - \frac{1}{2}gt^2$

where g = 9.8 m/s^2 is the acceleration of gravity. Substituting t = 2.04 s, we find

$y=(11.77)(2.04)-\frac{1}{2}(9.8)(2.04)^2=3.62 m$

The crossbar height is 3.05 m, so the difference is

$\Delta h = 3.62 - 3.05 =0.57 m$

So the ball passes 0.57 m above the crossbar.

8 0

3 years ago

How did private industries benefit from military expenditures?

Dafna11 [192]

Answer:

option A: the government paid private industries for goods and services.

Explanation:

8 0

2 years ago

) A 100-g ball falls from a window that is 12 m above ground level and experiences no significant air resistance as it falls. Wh

yawa3891 [41]

Answer:

Momentum = 1.534 kgm/s

Explanation:

Using the equations of motion, we can obtain the velocity of the ball as it hits the ground.

g = 9.8 m/s²

y = 12 m

u = initial velocity = 0 m/s, since the ball was released from rest

v = final velocity befor the ball hits the ground.

v² = u² + 2ay

v² = 0 + 2×9.8×12 = 235.2

v = 15.34 m/s

The momentum at any point is given as mass × velocity at that point

Mass = 100 g = 0.1 kg, velocity = 15.34 m/s

Momentum = 0.1 × 15.34 = 1.534 kgm/s

3 0

4 years ago