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prohojiy [21]
2 years ago
14

A bike rider does work to get to the top of a mountain. How could you use the law of conservation of energy to describe his trip

down the mountain? A. The rider uses energy to do work going up the mountain. B. This energy is converted to potential energy and then changes into kinetic energy on the way down. C. The bike rider's stored chemical energy allows him to move down the mountain, storing more and more energy as he goes. D.The bike rider gains electrical energy going up, which changes into electromagnetic energy on the way down.
Physics
1 answer:
netineya [11]2 years ago
5 0
The correct answer is (a.) The rider uses energy to do work going up the mountain. As the law of conservation of energy states that the total energy must be constant, therefore, the energy that is used to go up will be as it is to go down.
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Raghu studies in grade 6th. He wants a cricket bat to be made by a carpenter. He tells the
kvasek [131]

Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.

From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.

Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.

Learn more about length on:

brainly.com/question/25292087

8 0
2 years ago
What will be the ME of a machine that produces a 240 N work with a 300
Elden [556K]

Answer:

Efficiency = 80%

Explanation:

Given the following data;

Work output = 240 N

Work Input = 300 N

To find the mechanical efficiency of a machine;

Efficiency = \frac {Out-put \; work}{In-put \; work} * 100

Substituting into the equation, we have;

Efficiency = \frac {240}{300} * 100

Efficiency = 0.8 * 100

Efficiency = 80%

Therefore, the mechanical efficiency of the machine is 80 percent.

3 0
2 years ago
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
2 years ago
How much force is required to accelerate a 12 kg mass at 5 m/s 2
Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

4 0
2 years ago
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
2 years ago
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