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Andreyy89
3 years ago
9

A pile driver of mass 5 tonnes falls from a height of 10m onto a pile of mass 8 tonnes There is no rebound on impact as the pile

is driven 20cm into the ground. Determine the common Velocity of hammer and pile after impact.​
Physics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

5. View 31 suppliers of Drop Hammer Pile Driving Rigs on Suppliers. In very soft ... 2 m onto the top of a 140-kg pile, driving it 110 unit into the ground. ... Pilehire have drop hammers from 3ton to 30ton to suit all piling and post jobs ... The DELMAG diesel pile hammer is an extremely rugged and reliable impact hammer .

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KengaRu [80]

Answer:

Explanation:

You could try to say how helpful they are what they are and what they do

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2 years ago
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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
Mercury, Venus, Earth, and Mars are all ______ planets.
Pepsi [2]
<span>Mercury, Venus, Earth, and Mars are all ___inner___ planets. This is because they are all within the asteroid belt.</span>
8 0
3 years ago
A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck The puck travels for
Norma-Jean [14]

Answer:

Explanation:

Let's analyze the situation presented in order to know which answer is correct.

When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.

Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.

The correct answer is:

C)           Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck

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V = IR
I = V/R
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I = 0.6 amperes

Therefore, the current that is flowing through the wire supplied with 12 V and having a resistance of 20 
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