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3 years ago

15

An object has a mass of 1kg on Earth. What is its weight on the moon?

1 answer:

enyata [817]3 years ago

5 0

Answer:

the Mass of 1 kg object is same in Earth & Moon.

Explanation:

Weight, on the otherhand does change with location depends on the gravity. so the answer is : Weight of one kilo on the surface of moon is 1.622 N. A body is taken from the center of the Earth to the Moon.

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A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.

Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

$P = \frac{W}{t}$

$W= F*s$

Therefore,

$P=\frac{F*s}{t}$

Substitute the given values of force , displacement and time

$F = 124 N,s = 32.0 m,t = 8.0 s$

$P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W$

Thus the Power can be rounded off to the nearest value of 500 W

3 0

3 years ago

Read 2 more answers

What is the wavelength in air of red light from a helium neon laser?

Elza [17]

Answer:

632.8 nm is the wavelength (in air) of red light from a helium neon laser.

4 0

2 years ago

However, sometimes levers are designed to increase the input force needed to move an object. Which of the following is most like

Tems11 [23]

What are the answer choices, if there are any?

6 0

3 years ago

Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What

Sauron [17]

Answer:

0.426 L

Explanation:

Boyles law is expressed as p1v1=p2v2 where

P1 is first pressure, v1 is first volume

P2 is second pressure, v2 is second volume.

Given information

P1=96 kPa, v1=0.45 l

P2=101.3 kpa

Unknown is v2

Making v2 the subject from Boyle's law

$v2=\frac {p1v1}{p2}$

Substituting the given values then

$v2=\frac {96*0.45}{101.3}=0.4264560710760l\approx 0.426 l$

Therefore, the volume is approximately 0.426 L

6 0

3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

3 years ago