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astra-53 [7]
3 years ago
6

A fan powered by a 0.5 hp motor and delivers air at a rate of 85 m3 / min. Calculate the speed of the air moving by the fan if t

he air has a density of 1.18 kg / m3. Help me please
Physics
1 answer:
Elenna [48]3 years ago
4 0

s the following graph a bipartite graph or not, and why?

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Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses? A. specific heat B. conductiv
KengaRu [80]

Answer:

Efficiency

Explanation:

  • The amount of mechanical work an engine can do per unit of heat energy it uses is called its efficiency.
  • It is also defined as the output divided by the total electrical power consumed.
  • In terms of heat, efficiency of engine is given by :

\eta=1-\dfrac{Q_o}{Q_i}

Q_o\ and\ Q_i are output heat and input heat respectively.

Hence, the correct option is (d) "efficiency"

4 0
3 years ago
Why does a solid keep its shape
garri49 [273]

Answer:

Solids can hold their shape because their molecules are tightly packed together. Atoms and molecules in liquids and gases are bouncing and floating around, free to move where they want. The molecules in a solid are stuck in a specific structure or arrangement of atoms.

7 0
3 years ago
A circuit contains three light bulbs in parallel. After observation, a fourth light bulb is added, also in parallel. How does th
uranmaximum [27]
It has the same intensity.
6 0
2 years ago
Read 2 more answers
When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi
nydimaria [60]

Answer:

Correct option is A -  The force of kinetic friction is smaller than that of static friction, but Fg remains the same.

Explanation:

(A) The force of kinetic friction is smaller than that of static friction, but Fg remains the same

6 0
3 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
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