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2 years ago

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A ball is dropped from the top of a tall building. As the ball falls, the upward force of air rsistance becomes equal to the dow

nward pull of gravity. When these two forces become equal in magnitude the ball will: a) flatten due to the forces b) fall at a constant speed c) continue to speed up d) slow to a stop

1 answer:

iren [92.7K]2 years ago

5 0

Fall at a constant speed

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Pls help will mark Brainliest

BartSMP [9]

Answer: 100 units

Explanation:

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3 years ago

A box is being pulled to the right. What is the direction of the gravitational force?

lapo4ka [179]

The correct answer to the question is vertically downward i.e towards the centre of earth.

EXPLANATION:

As per the question, the box is pulled to the right.

Hence, the direction of the applied force is towards right.

We are asked to determine the direction of the gravitational force that acts on the body.

Before answering this question, first we gave to understand the gravitational force of earth.

Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre. It is equivalent to the weight of the body.

The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.

Hence, the direction of the gravitational force which acts on the box is vertically downward.

7 0

3 years ago

Read 2 more answers

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

2 years ago

Which formula can be used to find the X component of the resultant vector?

julia-pushkina [17]

Answer:

If you are simply looking for the X component then the most applicable formula from the choices given is Tx + Ux+ Vx. This means that you will add all x-components. For example: If a man walking along the x-axis walks 10 meters to the right, 5 back and 2 meters forward, what is the resultant vector?

4 0

2 years ago

An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

Marina86 [1]

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

4 0

2 years ago