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Verizon [17]
3 years ago
10

What is the potential energy, if abody of mass 250 kg is at a height of 30metre? ​

Physics
1 answer:
stepan [7]3 years ago
5 0
PE = (250) (30) (9.81) = 73575J
I used to use 9.81 as acceleration due to gravity. The answer might be different from what you expect
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488 J of work is done to a box which is moved across the floor for a distance of 8.9 m. What net force is required to act on the
kvv77 [185]
According to the formula
a = f \times d
Where a is work, f is force and d is the distance that box was moved over. And from that formula, you can get that f = a/d and that is 54.83N of force
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4 years ago
Why is it easier to open a door by holding it from it's edge​
Crazy boy [7]

Answer:

leverage

Explanation:

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3 0
2 years ago
a light-year is ____ A. the amount of time it takes light to get to the nearest star B. the distance light travels in a day C. t
Vinil7 [7]
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6 0
3 years ago
Read 2 more answers
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
4 years ago
A book is sliding along a desk top. Because the book is in motion ,you know that the forces acting on the book are A. Balanced B
Virty [35]
Just because the book is moving doesn't tell you anything about the forces on it, or even whether there ARE any.

Just look at Newton's first law of motion, and this time, let's try and THINK about it too. It says something to the effect that any object continues in constant, uniform MOTION ..... UNLESS acted on by an external force.
4 0
3 years ago
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