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3 years ago

Help fast please!!!!!

Physics

2 answers:

Svet_ta [14]3 years ago

8 0

Answer:

the answer is A

Hope tis helps:)

Explanation:

inysia [295]3 years ago

6 0

We're told that the planets have EQUAL MASS.

If that's true, then the strength of the gravitational forces between
each planet and the star depends only on the distance between
them ... the farther a planet is from the star, the smaller the
gravitational forces are IF we're talking about planets with
equal masses.

Planet-X is closer to the star, and Planet-Y is farther from it.
From this we know that the gravitational forces between the
star and Planet-X are greater, and the forces between the star
and Planet-Y are smaller.

'A' says this.

'B' is totally absurd, because it talks about gravity repelling things.

'C' says exactly the opposite for the two planets.

'D' says that distance doesn't matter. We know this is absurd,
simply because we're never pulled toward Jupiter in our daily life.

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Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

Given:

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

Assume:

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago

Where would a boat produce the highest concentration of carbon monoxide?

mr_godi [17]

A boat would produce the highest concentration of carbon monoxide in the exhaust system.

Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is slightly less dense than air. It is toxic to hemoglobic animals (both invertebrate and vertebrate, including humans) when encountered in concentrations above about 35 ppm.

5 0

3 years ago

Can someone please help me?

Alina [70]

Answer:

D would be it

Explanation:

cause none of the rest Makes sense to the book of the picture and I'm 100% sure =)

6 0

3 years ago

Your friend is wearing a red coat. When white light hits the coat, some light is reflected, and some is absorbed.

Vilka [71]

Answer:

Orange , yellow, green and blue

red coat absorbs all colors of visible light except red, so red light

is the only light left to bounce off of the coat toward our eyes.

4 0

2 years ago

A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a

GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

$T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}$

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

$T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}$

T = 7.83 X10⁻⁷ s

6 0

3 years ago