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Citrus2011 [14]
3 years ago
13

A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the

student to overcome gravity is most nearly
1.1 W
87.5 W
560 W
875 W
5600 W
Physics
1 answer:
nasty-shy [4]3 years ago
3 0

Answer:

The average power the student expended to overcome gravity is 560W, Watts is a units of work it is Joules /time or kg*m^{2}/ s^{3}

Explanation:

Weight = 700N

F= m*g*h \\m*g= 700N\\F= 700N*h\\F=700N*8m\\F= 5600 N*m \\F=5600 J

The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work

Work= \frac{5600 J}{10s} = \frac{5600 \frac{kg*m^{2} }{s^{2} } }{10 s}  \\Work= 560\frac{kg*m^{2} }{s^{3} } \\Work =560 W

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Describe the position of the sun, moon, and earth during a new moon and a full moon.
Gennadij [26K]

Answer:

* he new moon phase when the position is       Sun - Moon - Earth,

* have of the Full Moon when the position is     Sun - Earth - Moon,

*All the phases of the moon are governed by the movement of the Moon around the Earth.

Explanation:

In the solar system, the planets revolve around the sun, which is much more massive, in the case of the Earth it is more massive than its satellite, therefore the Moon revolves around the Earth in a period of approximately 28 days.

It is said that the moon is in the new moon phase when the position is Sun - Moon - Earth, so the moon cannot be seen

It is in the phase of the Full Moon when the position is

                  Sun - Earth - Moon, in this case the moon can be observed by the light reflected from it.

All the phases of the moon are governed by the movement of the Moon around the Earth.

8 0
3 years ago
The current through a 10 ohm resistor is 1.2 amperes.What is the potential difference across the resistor?
Natalija [7]

<u>Answer</u>: The potential difference across the resistor is 12 volts.

<u>Explanation:</u>

To calculate the potential difference cross the resistor, we use Ohm's Law. This law states that the potential difference across two wires is directly proportional to the current flowing through that wire.

Mathematically,

V\propto I\\V=IR

Where,

V = potential difference = ?V

I = Current flowing = 1.2 A

R = Resistor = 10\Omega

Putting values in above equation, we get:

V=1.2\times 10=12V

Hence, the potential difference across the resistor is 12 volts

6 0
3 years ago
A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th
fiasKO [112]
Let's use the mirror equation to solve the problem:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}
where f is the focal length of the mirror, d_o the distance of the object from the mirror, and d_i the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for d_i by using the numbers given in the text of the problem:
\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}
\frac{1}{d_i}= -\frac{7}{60 cm}
d_i = -8.6 cm
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
6 0
3 years ago
Read 2 more answers
When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels
Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

3 0
3 years ago
An object has a kinetic energy ke, and potential energy pe. it also has a rest energy e0. what is the object's total energy e?
Dmitriy789 [7]
Look that one up in you text book PG:678 that is if you got the same book as my friend<span />
7 0
3 years ago
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