All categories

3 years ago

What can be broken but never held?

Physics

2 answers:

Alex777 [14]3 years ago

6 0

Answer:

A promise :)

Explanation:

Hope this helps :)

geniusboy [140]3 years ago

4 0

Answer:a heart

Explanation:

You might be interested in

Which of these could NOT be used to find the mechanical advantage of an inclined plane?

34kurt

The "c) percent efficiency" could not be used to find the mechanical advantage of an inclined plane. There are two formulae that could be used to determine the mechanical advantage of an inclined plane which stated as MA = Length/rise and Wout=Win. MA is the mechanical advantage, Wout is the output force, Win is the input force, and "rise" is the height of the inclined plane<span>.</span>

8 0

3 years ago

How would the speed of Earth's orbit around the sun change if Earth's distance from the sun

gayaneshka [121]

If Earth’s distance from the sun increased by 4 times then the speed of Earth’s orbit around the sun would decrease by a factor of 2.

Answer: Option 3

<u>Explanation:</u>

From Newton’s second law of motion, it is known that any kind of external force acting on an object will be equal to object’s mass and acceleration exerted on the object. So in this case, the gravitational force between Earth and Sun will be equal to the mass (M) and acceleration exerted on Earth.

$\text {Gravitational force}=\text {M} \times \text {Acceleration of Earth around the orbit}$

It is known that,

$\text {Acceleration in orbital motion}=\frac{\left(\text {velocity at which the Earth rotates)}^{2}\right.}{\text { Distance of Earth from the Sun}}$

Thus, substituting this, we get,

$\text {Gravitational force}=\frac{\text {M} \times\left(\text {velocity at which the Earth rotates)}^{2}\right.}{\text { Distance of Earth from the sun}}$

Also, we know,

$\text {Gravitational force}=\frac{G \times M \text { of sun } \times \text {M of Earth}}{(\text {Distance of Earth from the Sun})^{2}}$

Then, comparing both the equation, we get the orbital velocity as,

$\text { Orbital velocity } v=\sqrt{\frac{G \times M \text { of Sun }}{\text { Distance of Earth from the Sun }}}=\sqrt{\frac{G M}{r}}$

So, here G is gravitational constant, M is the mass of Sun and r is the distance of separation of Earth from Sun.

If the distance of Earth from Sun increases by 4 times so r’ will be 4r. Thus the new orbital velocity v’ will be

$v^{\prime}=\sqrt{\frac{G M}{r^{\prime}}}=\sqrt{\frac{G M}{4 r}}=\frac{1}{2} \sqrt{\frac{G M}{r}}$

So,

$v^{\prime}=\frac{1}{2} v$

Thus, the orbital speed will be decreased by a factor of 2 when the distance of Earth from the Sun increased by 4 times.

7 0

3 years ago

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/

IrinaK [193]

Answer: (a) 1.065 N (b) 2.13 N

Explanation:

<h2>(a) average force exerted by the rain on the roof</h2><h2 />

According Newton's 2nd Law of Motion the force $F$ is defined as <u>the variation of linear momentum</u> $p$ <u>in time:</u>

$F=\frac{dp}{dt}$ (1)

Where the linear momentum is:

$p=mV$ (2) Being $m$ the mass and $V$ the velocity.

In the case of the rain drops, which initial velocity is $V_{i}=15m/s$ and final velocity is $V_{f}=0$ (we are told the drops come to rest after striking the roof). The momentum of the drops $p_{drops}$ is:

$p_{drops}=mV_{i}+mV_{f}$ (3)

If $V_{f}=0$ , then:

$p_{drops}=mV_{i}$ (4)

Now the force $F_{drops}$ exerted by the drops is:

$F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}$ (5)

$F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}$ (6)

At this point we know the mass of rain per second (mass rate) $\frac{dm}{dt}=0.071 kg/s$ and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

$F_{drops}=\frac{dm}{dt}V_{i}$ (7)

$F_{drops}=(0.071 kg/s)(15m/s)$ (8)

$F_{drops}=1.065kg.m/s^{2}=1.065N$ (9) This is the force exerted by the rain drops on the roof of the car.

<h2>(b) average force exerted by hailstones on the roof </h2><h2 />

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum $p_{hailstones}$ is: