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3 years ago

Which two types of energy does a book have as it falls to the floor

Physics

2 answers:

Murrr4er [49]3 years ago

8 0

Answer:

kinetic and potential energy

Explanation:

natita [175]3 years ago

3 0

☺️☺️☺️☺️zksoskksskksksksksksksksksks

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14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar

Zolol [24]

Answer:

$T= 11.0003$ s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

$\omega ^2 r =Angular\ acc$

Considering the outer ring,

$\omega ^2 r = 9.8$

$\omega ^2= \frac{9.8}{30}$

$\omega = \sqrt{\frac{9.8}{30}}$

$\omega= 0.571 rad/s$

Therefore solving for Period T

Generally the equation for solving Period T is mathematically given as

$T= \frac{2\pi}{\omega}$

$T= \frac{2\pi}{0.571 rad/s}$

$T= 11.0003$ s

5 0

3 years ago

Which statement is true of a concave lens?

natka813 [3]

It produces only virtual images is the answer

5 0

3 years ago

Read 2 more answers

A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with

romanna [79]

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = $30.1 \ m/s$

The velocity of red car = $45.4 \ m/s$

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

$t=\frac{45-30}{2.7}$ (∵ $time=\frac{distance}{time}$ )

$t=5.56 \ sec$

then,

The distance covered by trooper,

$t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2$

$=208.33 \ m$

The distance covered by red car,

= $45\times 5.56$

= $250.2 \ m$

Maximum distance = $250.2-208.33$

= $41.87 \ m$

6 0

3 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)