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chubhunter [2.5K]
3 years ago
5

surface is tipped to all principal planes of projection. Because it is not perpendicular to any projection plane, it cannot appe

ar on edge in any standard view. This surface always appears as a foreshortened surface in all three standard views.
Physics
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

Oblique surface

Explanation:

Surfaces or lines situated at an angle in relation to all the main projection planes are oblique. An oblique surface never appears in any of the main perspectives as a line or true size area. It's Tipped to all main projection planes. doesn't quite appear on the edge of every standard view. doesn't quite show up in any view in true size.it has always been foreshortened.

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A 0.155 kg arrow is shot upward
solniwko [45]

Answer:

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

 it can be derived using the expression below:

 

   K.E  = \frac{1}{2}  m v²

m is the mass

v is the velocity

 Solve for K.E;

    K.E  =  \frac{1}{2}  x 0.155 x 31.4 = 2.43J

3 0
3 years ago
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What is the part of the cell that stores food?
Alenkasestr [34]
Part of the cell that stores food is called vacuole
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
Explain how the thermal energy of an isolated system changes with time if the mechanical energy of that system is constant.​
Naddika [18.5K]

Answer:

Thermal energy of an isolated system changes with time If the mechanical energy of that system is constant according to the first law of thermodynamics, which states that thermal energy of an isolated system can still change as long as the total energy of that system does not change.

Explanation:

6 0
3 years ago
How many joules of work are done on an object when a force of 10 N pushes it 5 m?
zhenek [66]

Answer:

option C

Explanation:

given,                            

Force on the object = 10 N

distance of push = 5 m

Work done = ?              

we know,              

work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0
3 years ago
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