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tangare [24]
3 years ago
5

What is relative density​

Physics
1 answer:
Alja [10]3 years ago
4 0

Answer:

Relative density, or specific gravity, is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material.

Explanation:

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The transfer of energy when particles of a fluid move from one place to another is called
siniylev [52]

Answer:

Answer below!!!!

Explanation:

Convection is the transfer of thermal energy by particles moving through a fluid.

Hope I Helped!!!

;)

5 0
3 years ago
Record your model values in the table below
AlladinOne [14]
There not enough inform here to answer this question
8 0
4 years ago
At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 54.0 N at the s
butalik [34]

Answer:

m=5.51Kg

Explanation:

The weight of an object on Earth is given by W=mg, so we can calculate its mass by doing m=W/g, which for our values is:

m=(54N)/(9.8m/s^2)=5.51Kg

<em>Nothing is being asked</em> about Io but if one wanted to know the weight <em>W'</em> of the watermelon there one just have to do:

W'=ma=(5.51Kg)(1.81m/s^2)=9.97N

4 0
3 years ago
A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of
Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}

Where, \beta=\dfrac{c}{v}

\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}

\lambda_{o}=115\ nm

Hence, The observer detects light of wavelength is 115 nm.

8 0
4 years ago
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.64 10-2 kg/s. the density of the gasoline is 735
Irina18 [472]
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m

Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²

Let v =  speed of pumping the gasoline, m/s
Then the mass flow rate is 
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s

The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s

Answer:  2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
7 0
3 years ago
Read 2 more answers
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