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3 years ago

Which statement describes how work and power are similar?

2 answers:

6 0

The statement that describes how work and power are similar is D. you must know time and energy to calculate both.
I am not completely sure though, so I hope this helps. :)

sertanlavr [38]3 years ago

6 0

Which statement describes how work and power are similar?

Answer: Out of all the options shown above the one that best describes how work and power are similar is answer choice A) You must know force and distance to calculate both. The reason being that work equals force X distance and power equals work/time.

I hope it helps, Regards.

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Define mixture, heterogeneous, homogeneous, solution, colloid, suspension, solvent, solute, saturation.( please don't answer the

oksano4ka [1.4K]

Answer:

1) a substance made by mixing other substances together.

2) diverse in character or content.

3) of the same kind; alike.

4) a means of solving a problem or dealing with a difficult situation.

5) a homogeneous noncrystalline substance consisting of large molecules or ultramicroscopic particles of one substance dispersed through a second substance. Colloids include gels, sols, and emulsions; the particles do not settle, and cannot be separated out by ordinary filtering or centrifuging like those in a suspension.

4 0

3 years ago

Read 2 more answers

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$

We also know the total time of flight,

t = 7.82 s

Therefore, we can find the horizontal distance travelled by the sack:

$d=v_x t = (60)(7.82)=469.2 m$

So, the sack will land 469.2 m from the original point below the airplane.

Learn more about free fall and projectile motion:

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7 0

3 years ago

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

3 years ago

Seja uma carga Q=1.2×10(-8)C no vácuo, distando 40 cm de um ponto M e 50 cm de um ponto N . Determine os pontencias no ponto M e

Len [333]

<span>Americano, por favor?</span>

4 0

4 years ago

How long does it take for a 3.5 kW electric water heater to heat 40 kg of water? from 20 ° C to 75 ° C? The specific heat capaci

iren [92.7K]

Answer:

2633.7 s

Explanation:

From the question,

Heat lost by the water heater = Heat gained by the water

Applying,

P = cm(t₂-t₁)/t.................. Equation 1

Where P = power of the heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature, t = time

make t the subject of the equation

t = cm(t₂-t₁)/P.............. Equation 2

From the question,

Given: c = 4190 J/kgK, P = 3.5 kW = 3500 W, m = 40 kg, t₁ = 20°C, t₂ = 75°C

Substitute these values into equation 2

t = 4190×40(75-20)/3500

t = 9218000/3500

t = 2633.7 s

4 0

3 years ago