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3 years ago

11

What is the amount of "stuff in an object?

2 answers:

ziro4ka [17]3 years ago

7 0

mass is the amount of "stuff in an object

Andrei [34K]3 years ago

4 0

Good question. The amount of 'stuff' in an object is it's mass.

A fundamental distinction we learn in physics is the difference between mass and weight. If we were in deep space, away from any very large objects of mass (like a planet), we would be 'weightless' e.g. not feel the effects of gravity, but we would not be 'massless'. Our mass doesn't change based upon our proximity to large objects (gravitational attraction), but the sense of weight does.

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A car engine supplies 2.0 x 103 joules of energy during the 10. seconds it takes to accelerate the car along a horizontal surfac

andrew11 [14]

Answer: 2. 2.0*10^2 W

Explanation:

Power = Work/Time

Power = (2.0*10^3) Joules/10 seconds

Power = 2.0*10^2 Watts

7 0

2 years ago

Please help At which point is the potential energy the greatest ?

4vir4ik [10]

Answer: 3

Explanation:

5 0

2 years ago

A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the

Tpy6a [65]

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

$PV=nRT$

$P=\dfrac{WRT}{VM}$

$P=\dfrac{\rho RT}{M}$

$T=\dfrac{PM}{\rho R}$

Put the value into the formula

$T=\dfrac{2\times28}{1.7\times0.0821}$

$T=401.2\ K$

We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

$V_{rms}=\sqrt{\dfrac{3RT}{M}}$

$V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}$

$V_{rms}=1.87\ m/s$

Hence, The speed of nitrogen molecule is 1.87 m/s.

5 0

3 years ago

At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of

photoshop1234 [79]

The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.

So recoil velocity of cannon v2 is given by,

v2 = -(m1÷m2)v1

v2 = -(100÷500)15

v2 = -5 m/s

where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.

Hence, option (4)5.0 m/s is the correct answer.

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Rudik [331]

I think it is true I think

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3 years ago

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