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Nesterboy [21]
2 years ago
8

Biotic factors and abiotic factor temperature

Physics
1 answer:
ddd [48]2 years ago
5 0
Biotic is the living factors while abiotic is nonliving such as a rock
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a concrete cube of side 0.50 m and uniform density 2.0 x 103 kg m–3 is lifted 3.0 m vertically by a crane. what is the change in
AlladinOne [14]

Answer:

Change in potential energy = 7350 Joules

Explanation:

It is given that,

Side of cube, a = 0.5 m

Density of cube, d=2\times 10^3\ kg/m^3

The cube is lifted vertically by a crane to a height of 3 m

We know that, density d=\dfrac{m}{V}

So, m = d × V  (V = volume of cube = a³)

m=2\times 10^3\ kg/m^3\times (0.5\ m)^3

m = 250 kg

We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.

Potential energy at height h is given by :

PE=mgh

PE = 250 kg × 9.8 m/s² ×3 m

PE = 7350 Joules

So, change in potential energy of the cube is 7350 Joules.

5 0
3 years ago
Read 2 more answers
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
An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish
VikaD [51]
The answer is 7.1 I just took the test
8 0
3 years ago
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7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist
luda_lava [24]

Answer:  5.05 m

Explanation:

4 0
3 years ago
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Which of the following is strenuous?
Arada [10]

Answer:

I think it c

Explanation:

7 0
3 years ago
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