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2 years ago

What to do if one day your partner wants you but the next day are unsure about you should I try to make it work or leave ..

1 answer:

6 0

Leave them, listen you are better than being with someone who is unsure. If you are 100% in this they need to match the energy. The purpose of dating is going into marriage eventually and imagine if when you get married they are like “wait I don’t know actually”. True me dump them and don’t look back

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A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

charle [14.2K]

Answer:

She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

$77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

So she will cover 77 m in 2.85 seconds

Now considering vertical motion, up direction as positive

Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 $m/s^2$ , time = 2.85

Substituting

$s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

So she will make the jump.

6 0

2 years ago

How long does it take to fall from 5000 feet?

WINSTONCH [101]

It depends on your weight, your hieght, and how fast you are falling

4 0

3 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

A 4 kg car with frictionless wheels is on a ramp that makes a 25° angle with the horizontal. The car is connected to a 1 kg mass

stellarik [79]

Refer to the diagram shown below.

W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N

The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N

The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²

Answer: 1.69 m/s² (nearest hundredth)