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olganol [36]
3 years ago
9

Write the similarity between the inclined plane and a screw?​

Physics
1 answer:
lord [1]3 years ago
7 0

An inclined plane makes work easier by breaking an upward or downward movement into smaller increments.

A screw is like an inclined plane wrapped around a cylinder. It turns a small rotational force into a larger forward driving force

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If an inclined plane is 6 m long and 2 m high, what is its mechanical advantage?
choli [55]

Answer:

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3 years ago
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A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft
Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

m_1 v_1 = m_2 v_2

m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s

Using the equation to find v_1,

v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475

Using the conservation of energy equation, we have,

KE= \frac{1}{2}m*v^2

KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J

KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J

Total KE= 92077.75+13425530=92708.9J

Now this energy over the cannonball

KE=\frac{1}{2}m*v_2^2

92708.9=\frac{1}{2}15.5v_2^2

V_2 = 109.37m/s

The gain in velocity is 0.37m/s

4 0
3 years ago
If a layer was deposited but does not appear in the rock record, what has occured
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I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. <span>If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.</span>
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Which statement describes something that should always be done at the end of a calculation?
Reika [66]

Answer:

C) Check that the numerical answer is reasonable and the units are what you expected.

Explanation:

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3 years ago
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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
mash [69]

a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

T=\frac{2\pi (R+h)}{v}

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0
3 years ago
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