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4 years ago

It is raining outside, so a student lifts their textbook vertically 1.2 meters above

Physics

1 answer:

Nadusha1986 [10]4 years ago

3 0

Answer:

Work done is 50.4J.

Explanation:

The work $W$ done by the student in lifting the book 1.2 meters vertically is

$W =42N*1.2m$

$\boxed{W = 50.4J}$

When the student begins to walk horizontally, the net force on the book is zero <em>(gravity is pulling it down, the student is lifting it up)</em>. Thus, although the student is displacing the book horizontally, but because no force is exerted on it, the work done is zero.

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A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

4 0

3 years ago

Who much force is needed to accelerate a 68 kilogram-skier at a Rate of 1.2 m/sec

liberstina [14]

If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2

7 0

3 years ago

Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp

Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

$a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*t+\frac{1}{2} *2*t^{2}$

$t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s$

Check

$t_{2}=8s$

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}$

$x_{f}=0-6*+\frac{1}{2} *2*8^{2}$

$x_{f}=-48+64\\x_{f}=16$

5 0

4 years ago

The difference between the frequency fff and the frequency ωωomega is that fff is measured in cycles per second or hertz (abbrev

Kisachek [45]

Answer:

Radians

Explanation:

The angular speed is a measure of the rotation speed of a body. It is defined as the angle rotated by a unit of time. Thus, It refers to the angular displacement per unit time and is designated by the Greek letter $\omega$ . Its unit in the International System is radian per second (rad / s).

6 0

3 years ago

TRUE OR FALSE??

Stels [109]

Answer:

false, true, true, true, false

Explanation:

8 0

3 years ago