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Dmitrij [34]
3 years ago
15

If an object has a mass of 20 grams and a volume of 40 cm3, what is its density in g/cm3?

Physics
1 answer:
sattari [20]3 years ago
3 0
Density = Mass/Volume 
So, given mass = 20 g and volume = 40 cm^3 
By substituting in above equation, Density = 20/40 = 0.5 g/cm^3
Hope it helps.

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Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth
Semmy [17]

Answer:

\lambda=6.83\times 10^{-5}\ m

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

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So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m

So, the wavelength of the infrared radiation is 6.83\times 10^{-5}\ m.

5 0
2 years ago
You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55
ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

v=v_0+at\\\\v^2=v_0^2+2ad

Dividing the second equation by the first one, we obtain:

v=\frac{v_0^2+2ad}{v_0+at}

And, since v_0=0, then:

v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2

So the acceleration is 3.30m/s^2.

7 0
3 years ago
Which of the following is an example of Class 3 lever system
insens350 [35]
If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.
6 0
2 years ago
Read 2 more answers
A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875
mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

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= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

5 0
2 years ago
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What is the equivalent resistance between points A and C if R1=1430, R2=1350, R3=1100, R4=1350, and R5=1150.
Marianna [84]

R1 + R4 = 1430 + 1350 = 2780 = R14    series combination of R1 & R4

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7 0
3 years ago
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