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2 years ago

.....i need help with another......

Physics

2 answers:

BartSMP [9]2 years ago

7 0

Your answer to this is B

Masja [62]2 years ago

4 0

The screw is really a twisted inclined plane. The answer is D i know it !! You're welcome :)

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What is a non contact force that attracts all objects to the centre of the earth

beks73 [17]

Answer:

Gravity. The weight-force or weight of an object is the force because of Gravity, which acts on the object attracting it towards the centre of the earth.

Hope this helps, 

 x

4 0

2 years ago

Neo and the agent stand in different locations and the length of the line between them (their distance apart) is a measurement o

scZoUnD [109]

I think their distance is a measurement of : B. space in two dimension

In two-dimensional space, both directions located in the same plane , and the distance in locations only separated by width and length (there is no volume in this model)

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2 years ago

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A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis

romanna [79]

Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

$v=\omega r$

where $\omega$ is the angular velocity and $r$ is the radius.

$\omega=\dfrac{v}{r}$

The edge has a linear velocity of 10 m/s and the radius at the edge is 5 m.

$\omega=\dfrac{10 \text{ m/s}}{5 \text{ m}} = 2 \text{ rad/s}$

8 0

3 years ago

Look at the diagram At which position on the diagram, does the ball have the least kinetic energy and the greatest potential ene

Anestetic [448]

I think it’s point C

4 0

2 years ago

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An object of mass m is traveling around a circle of radius v. What happens to the centripetal force if the radius is cut in half

german

Answer:

Centripetal force doubles

Explanation:

Since centripetal force is given by

$F=\frac {mv^{2}}{r}$

Where F is the centripetal force, m is the mass, v is the centripetal velocity and r is circle radius.

When the radius is reduced to half then the centripetal force will be

$F=\frac {mv^{2}}{0.5r}=\frac {2mv^{2}}{r}$

Evidently, the centripetal force is doubled compared to the initial.

Note that while the question gives letter v for radius, I used v for velocity and r for radius since these are the standard letters for abbreviating formula of centripetal force.

7 0

2 years ago