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ValentinkaMS [17]

1 year ago

12

A bodybuilder lifts a 10 N weight a distance of 2.5 m. How much energy has the weight gained?

1 answer:

dolphi86 [110]1 year ago

3 0

Answer:

25 N

Explanation:

2.5 * 10 = 25

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Stella is driving down a steep hill. She should keep her car __________ to help _________. a. light, it speed up in a higher b.

frez [133]

Answer:

<em>The Answer is both B and C, </em><em>since it has same options from the question given. Gear, slow her vehicle in a lower</em>

Explanation:

<em>The use of a lower gear in a vehicle helps a person to control their speed limits, when approaching a hill. it also saves the brakes too, using the brakes down a hill can overheat the gear and causes brake failures</em>

<em>By changing in into a lower gear and also letting the engine to do the brake work in a vehicle, the engine will absorb a force and slow the vehicle down, but in most cases brakes can be applied but with lesser pressure.</em>

<em>In this case Stella need to slow down by applying her lower gear down a hill to avoid accidents on the road, by controlling her speed limits and for safety precaution</em>

3 0

2 years ago

Read 2 more answers

Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance (in centim

Andrej [43]

Answer:Your ear is capable of differentiating sounds that arrive at the ear just 1.00 ms apart. What is the minimum distance between two speakers that produce sounds that arrive at noticeably different times on a day when the speed of sound is 340 m/s

Explanation:

5 0

2 years ago

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

lana66690 [7]

Answer:

E) I = 18.4 N.s

Explanation:

For this exercise let's use momentum momentum

I = Δp = $p_{f}$ - p₀

The energy of the stone is only kinetic

K = ½ m v²

The initial energy is Ko and the final is 70% Ko

$K_{f}$ = 0.70 K₀

energy equation

$K_{f}$ = 0.7 ½ m v₀²

You can also write

$K_{f}$ = ½ m vf²

½ m vf² = ½ m (0.7 v₀²)

$v_{f}$ = v₀ √ 0.7

Now we can calculate and imposed

I = m (-vo √0.7) - m vo

I = m vo (1 +√0.7

I = 0.5000 20.0 (1.8366)

I = 18.4 N.s

5 0

2 years ago

Marcia is given an incomplete chemical equation that includes the number of nitrogen atoms present in the products of the reacti

Semmy [17]

<span>If the products contain 3 nitrogen atoms, then so did the reactants since overall mass is conserved in a chemical reaction. </span>

8 0

2 years ago

Read 2 more answers

In this example we will determine what type of lens should be used to correct the vision of a hyperopic eye. Assume that the nea

masha68 [24]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a

The lens to be used is a positive converging lens with focal length of f = 33.33 cm

b

The lens to be used is a positive converging lens with focal length of f = 27.3 cm

Explanation:

a

From the question and the diagram we are told that

The image distance is $v = - 100cm$ in front of the eye

The object distance is $u = -25cm$

According to the Lens formula the focal length of the lens to be used to correct the hyperopic eye is

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{-100} - \frac{1}{(-25)} = \frac{1}{f}$

$\frac{1}{f} = \frac{1}{25} - \frac{1}{100}$

$\frac{1}{f} = \frac{75}{25 * 100}$

$f = \frac{100}{3}$

$f = 33.33cm$

Since the focal length is positive it means that the lens to use is a positive

converging lens

b

From the question we are told

The image distance is $v = -300cm$ in front of the eye

The object distance is $u = -25cm$

According to the Lens formula the focal length of the lens to be used to correct the hyperopic eye is

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{-300} - \frac{1}{(-25)} = \frac{1}{f}$

$\frac{1}{f} = \frac{1}{25} - \frac{1}{300}$

$\frac{1}{f} = \frac{275}{25 * 300}$

$f = 27.3$

Since the focal length is positive it means that the lens to use is a converging lens

4 0

2 years ago