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never [62]
2 years ago
12

Air only has a density of 1.3kg/m². What is the mass of3m^3 of air?​

Physics
1 answer:
Katarina [22]2 years ago
8 0

Answer:

m = 3.9 [m³]

Explanation:

We must remember that the density of the substance is defined as the relationship between mass over volume. That is expressed in the following equation:

Ro = m/V

where:

Ro = density = 1.3 [kg/m³]

m = mass [kg]

V = volume = 3 [m³]

Now replacing the values.

m =Ro*V\\m = 1.3*3\\\\m=3.9[m^{3}]

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3 years ago
Explain how work is related to energy
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Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work.


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3 years ago
The temperature of a balloon is lowered with liquid nitrogen. The balloon appears to deflate because the A. nitrogen has burst t
Elina [12.6K]

The temperature of a balloon is lowered with liquid nitrogen. The balloon appears to deflate because the gas in the balloon lost energy and condensed. The answer is letter D.

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3 years ago
A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

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3 years ago
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Genrish500 [490]

Answer:

D

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