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Ronch [10]
4 years ago
5

How does friction affect a machine efficiency?

Physics
1 answer:
MariettaO [177]4 years ago
5 0
It wears down its components causing it to not work as effectively, it also creates heat which can cause(according on the machine) it to burn out. Another Immediate problem is that it can slow the machine causing it to lose efficiency.
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A revolutionary war cannon, with a mass of 2090 kg, fires a 16.7 kg ball horizontally. The cannonball has a speed of 113 m/s aft
OLEGan [10]

Answer:

0.90291 m/s

0.45055 m/s

Explanation:

m_1 = Mass of canon = 2090 kg

m_2 = Mass of ball = 16.7 kg

v_1 = Velocity of canon

v_2 = Velocity of ball = 113 m/s

In this system the momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{16.7\times 113}{2090}\\\Rightarrow v_1=0.90291\ m/s

The velocity of the cannon is 0.90291 m/s

Applying energy conservation

\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2=\dfrac{1}{2}m_2v^2\\\Rightarrow m_1v_1^2+m_2v_2^2=m_2v^2\\\Rightarrow v_2=\sqrt{\dfrac{m_1v_1^2+m_2v_2^2}{m_2}}\\\Rightarrow v_2=\sqrt{\dfrac{2090\times 0.90291^2+16.7\times 113^2}{16.7}}\\\Rightarrow v_2=113.45055\ m/s

The ball would travel 113.45055-113 = 0.45055 m/s faster

6 0
4 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
According to current scientific estimates, when did the big bang occur?.
Bess [88]

Answer:

13.8 billion years ago

Explanation:

3 0
3 years ago
Can someone help me pls??
Jet001 [13]

Answer:

newton's first law (sorry its really late i know you prob don't need it anymore

Explanation:

3 0
3 years ago
What is the base matterial for plastic wrap
ValentinkaMS [17]
Plastic wrap basic material is <span>polyvinyl chloride or PVC.

Hope this helps, 

kwrob</span>
4 0
3 years ago
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