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salantis [7]
3 years ago
12

What is the power of a student that has done a work of 10 joules in 10 seconds​

Physics
1 answer:
Alla [95]3 years ago
8 0

Answer:

1 Watt

Explanation:

P=W/t

P=10/10

P=1 Watt

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A small cart travels 20 meters to the right in 10 seconds calculate the velocity
stira [4]
The velocity would be 10
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3 years ago
How do gravitational and electic force compare
velikii [3]

Electric Forces. ... Just like objects that have mass exert gravitational forces on each other, objects that are charged will also exert electric forces on each other. The electric force is directly proportional to the charge of the two objects and inversely proportional to the distance between them squared.

3 0
3 years ago
Give an example of free fall​
Elena-2011 [213]

Answer:

A stone that is dropped down into an empty well

3 0
2 years ago
2.
Alex_Xolod [135]

Answer:

The work done on the sled by friction, W = - 4593.75 J

Explanation:

Given data,

The combined mass of sled and the boy, m = 75 kg

The displacement of the boy, S = 25 m

The coefficient of the friction, u = 0.25

The frictional force acting on the boy,

                  <em>F = u η</em>

Where,

                        η - is the normal force acting on the boy (mg)

Substituting the values,

                   F = 0.25 x 75 x 9.8

                      = 183.75 N

Since the direction of the frictional force is against the direction of motion

                      F = - 183.75 N

The work done on the sled by friction,

                         W = F x S

                             = - 183.75 x 25

                             = - 4593.75 J

Hence, the work done on the sled by friction, W = - 4593.75 J

5 0
3 years ago
A lead fishing weight of mass 0.20 kg is tied to a fishing line that is 0.50 m long. The weight is then whirled in a vertical ci
solong [7]

Answer:

Following are the solution to this question

Explanation:

please find the complete question in the attached file.

In point a:

The answer is "bottom".

In point b:

Using formula:

T= mg + \frac{m V^2}{2}

\to 100= 0.2 \times 9.81 + \frac{0.2 \times V^2}{0.5}

\to 100= 1.962+ \frac{0.2 + V^2}{0.5}\\\\\to  100- 1.962= \frac{0.2 + V^2}{0.5}\\\\\to  98.038= \frac{0.2 + V^2}{0.5}\\\\\to  49.019=0.2+V^2\\\\\to  48.819=V^2\\\\ \to  6.987 \ \frac{m}{s}

6 0
3 years ago
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