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AleksAgata [21]
3 years ago
5

Which of the following is an example of an exerting force?

Physics
1 answer:
Firlakuza [10]3 years ago
4 0
A carpenter hammering a nail
Explanation:
This is the only choice that gives you an object exerting a force ( the carpenter/hammer ) and one that has a force exerted on it ( the bail )
All of the rest would be related to acceleration and speed
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At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is
Helen [10]

Answer:

The thrown rock will strike the ground 2.42s earlier than the dropped rock.

Explanation:

<u>Known Data</u>

  • y_{i}=300m
  • y_{f}=0m
  • v_{iD}=0m/s
  • v_{iT}=-29m/s, it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}, then clearing for t_{D}.

t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s

<u>Time of the Thrown Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}, then, 0=-4.9t_{T}^{2}-29t_{T}+300, as it is a second-grade polynomial, we find that its positive root is t_{T}=5.4s

Finally, we can find how much earlier does the thrown rock strike the ground, so t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s

6 0
4 years ago
Answer this please nowww
Viefleur [7K]
(3) The frictional force exerted by the floor on the box
4 0
3 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
A rain cloud contains 5.32 × 107 kg of water vapor. The acceleration of gravity is 9.81 m/s 2 . How long would it take for a 2.0
Alja [10]

Answer:

20.85 years

Explanation:

2.61 km = 2610 m

2.07 kW = 2070 W

First we need to calculate the potential energy required to take m = 5.32 * 10^7 kg of rain cloud to an altitude of 2610 m is

E = mgh = 5.32 * 10^7 * 9.81 * 2610 = 1.4*10^{12}J

With a P = 2070 W power pump, this can be done within a time frame of

t = E/P = \frac{1.4*10^{12}}{2070} = 658037739 s

or 658037739/(60*60) = 182788 hours or 182788 / 24 = 7616 days or 7616 / 365.25 = 20.85 years

4 0
4 years ago
1. why did aristarchus choose the time of a half (quarter) moon to make his measurements for calculating the earth-sun distance?
Stells [14]

In order to make his measurements for determining the Earth-Sun distance, Aristarchus waited for the Moon's phase to be exactly half full while the Sun was still visible in the sky. For this reason, he chose the time of a half (quarter) moon.

<h3 /><h3>How did Aristarchus calculate the distance to the Sun?</h3>

It was now possible for another Greek astronomer, Aristarchus, to attempt to determine the Earth's distance from the Sun after learning the distance to the Moon. Aristarchus discovered that the Moon, the Earth, and the Sun formed a right triangle when they were all equally illuminated. Now that he was aware of the distance between the Earth and the Moon, all he needed to know to calculate the Sun's distance was the current angle between the Moon and the Sun. It was a wonderful argument that was weakened by scant evidence. Aristarchus calculated this angle to be 87 degrees using only his eyes, which was not far off from the actual number of 89.83 degrees. But when there are significant distances involved, even slight inaccuracies might suddenly become significant. His outcome was more than a thousand times off.

To know more about how Aristarchus calculate the distance to the Sun, visit:

brainly.com/question/26241069

#SPJ4

7 0
2 years ago
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