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vampirchik [111]
3 years ago
14

A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of f

riction between the box and the floor is μ. At what rate does the child do work on the box?
Physics
1 answer:
dexar [7]3 years ago
4 0

Answer:

Explanation:

Given

mass of box is m

speed of box is v

coefficient of friction \mu =\frac{1}{4}=0.25

Box is moving at a constant speed i.e. Net Force on box is zero

i.e. Frictional Force and Force Applied are Equal

F=f_r=\mu mg

The Rate at which power is generated is

P=F\cdot v

P=\mu mg\cdot v

P=0.25mgv

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Which types of heat transfer involve heat flow from hot objects to colder objects? A. convection, conduction, and radiation B. c
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While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

4 0
4 years ago
Which vector has an x-component with a length of 3?<br> А. С.<br> B. d<br> C. a<br> D. b
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Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
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The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

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