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3 years ago

A wheel and axle has another simple machine in it. what is it

Physics

1 answer:

prisoha [69]3 years ago

7 0

Answer:

The wheel and axle is another simple machine. It uses a wheel with a rod attached in the middle as an axle to help it to lift or move loads. In some cases this machine works like a lever to multiply force (like with a doorknob or a fishing reel).

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A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont

Georgia [21]

Answer:

$141.30grams$

Explanation:

First, denote our known values;

$J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?$

Mass is impulse divided by change in velocity:

$m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs$

Hence, the mass of the ball is 141.30grams

3 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

A tectonic plate is part of the Earth’s lithospherehttp://brainly.com/question/add?entry=1642&task_content=A%20tectonic%20pl

lord [1]

Plate tectonicsis a scientific theory that describes the large-scale motion of Earth's lithosphere. This theoretical model builds on the concept of continental drift which was developed during the first few decades of the 20th century. The geoscientific community accepted plate-tectonic theory after seafloor spreading was validated in the late 1950s and early 1960s.The lithosphere, which is the rigid outermost shell of a planet (the crust and upper mantle), is broken up into tectonic plates.

3 0

3 years ago

How do you calculate the power in a circuit or device if given the current and the resistance?

Drupady [299]

Answer:

you use the Ohms law so to find the voltage you would need to multiply the current by the resistance which gives you the power.

V- voltage

I- current

R- resistance

V= I×R

5 0

3 years ago

Urgent need help 100 points

Sati [7]

Answer: The velocity with which the sand throw is 24.2 m/s.

Explanation:

Explanation:

acceleration due to gravity, a = 3.9 m/s2

height, h = 75 m

final velocity, v = 0

Let the initial velocity at the time of throw is u.

Use third equation of motion

The velocity with which the sand throw is 24.2 m/s.

7 0

2 years ago