Physics help please

a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d

USPshnik [31]

Answer:

The time is $t = 2.595 \ s$

The speed is $v = 25.43 \ m/s$

Explanation:

From the question we are told that

The height of the cliff is $h = 33 \ m$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} gt^2$

before the jump the persons initial velocity is u = 0 m/s

So

$33 = 0 * t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.595 \ s$

Generally from kinematic equation

$v= u + gt$

=> $v= 0 + 9.8 * 2.595$

=> $v = 25.43 \ m/s$

3 0

2 years ago

A ________ is a government-regulated maximum price for goods.

Simora [160]

D. price ceiling
This is a government regulation that establishes a maximum price for a particular good.

8 0

2 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

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

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

2 years ago

Care este pozitia unui mobil cand t(s)=14 si x(m)=12?

mario62 [17]

name shrawan sha

DtD_LpOZ3Xm_0ff

3 0

2 years ago

What is the definition of Mutualism

noname [10]

Mutualism is a long-term relationship where two organisms interact in such a way that both of them benefits from that relationship.

For example, there is a relationship between a bird called "oxpecker", and a rhino. While the bird eats the harmful bugs (eg. tick) on the rhino's skin and relieves its hunger; the rhino gets rid of the bugs that harm it.

5 0

3 years ago

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