All categories

3 years ago

A 5 kg rock is dropped down a vertical mine shaft. How long does it take to reach the bottom, 79 meters below?

Physics

1 answer:

likoan [24]3 years ago

7 0

Answer:

The time for the rock to reach the bottom is 4.02 seconds.

Explanation:

Given;

mass of the rock, m = 5 kg

height of the rock fall, h = 79 meters

The time to drop to the given height is given by;

$t = \sqrt{\frac{2h}{g} }$

where;

t is the time to fall to the bottom

g is acceleration due to gravity = 9.8 m/s²

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s$

Therefore, the time for the rock to reach the bottom is 4.02 seconds.

You might be interested in

The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

Equations

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

6 0

3 years ago

Does the refraction of light make a swimming pool seem deeper or shallower? explain your answer.

larisa [96]

The refraction of light makes a swimming pool seem shallower.

The swimming pool seems shallower because the rays of light coming from the bottom of the pool do not come with a straight path. The path of light is straight as long as it is in the water.

When lights come out of the water into the air it bents downwards. This bending is called refraction.

Refraction forms a virtual image of the pool and it seems shallower than it actually is to the observer. This only happens when light travels from one transparent medium into another having lower density.

If you need to learn more about why a swimming pool appears shallower, click here

https://brainly.in/question/7136803?referrer=searchResults

#SPJ4

6 0

1 year ago

Why does wave height increase in shallow water??

Vitek1552 [10]

As the water russhes toward the shore, it rises because it is pushing against it.

3 0

3 years ago

Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?

erastova [34]

I am pretty sure it is a cold front.

5 0

3 years ago

Read 2 more answers

If a pebble is being transported in a stream by rolling, how does the velocity of it compare to the velocity of the stream?

AleksandrR [38]

Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.

3 0

3 years ago