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3 years ago

If you push on the floor with 5 N for 15 seconds, how much work is done?

Physics

2 answers:

Maslowich3 years ago

4 0

Answer: It would be 75

Explanation: If your pushing on the floor at a rate of 5 for 15 seconds you would multiply the numbers and get 75

Schach [20]3 years ago

4 0

15*5=75
The amount of work that is done is 75.

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If we put negative charge between two similar positive charges then what is it's equilibrium? And how?

Gnesinka [82]

Your question has been heard loud and clear.

Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.

Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.

Thank you

5 0

3 years ago

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Could I please get some help on this question I don’t understand .

Oksana_A [137]

Answer:

12.5 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 8 m

Final velocity (v) at 8 m above the lowest point =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 8)

v² = 0 + 156.8

v² = 156.8

Take the square root of both side

v = √156.8

v = 12.5 m/s

Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.

5 0

3 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

3 years ago

Alexis is a scientist who is studying solid-state physics. Which activity would she most likely do as a part of this research? O

Ann [662]

Answer:

pretty sure its studying the atomic structure of a solid carbon dioxide. so c

Explanation:

8 0

2 years ago

8. Before leaving the ground an airplane traveling with constant acceleration makes a run on the

Effectus [21]

Answer:

$\color{Blue}\huge\boxed{Answer}$

<h3>B. Speed at which it leaves the ground</h3>

8 0

2 years ago

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