Please help :<

The gravitational force between two asteroids is 1,000,000 n. what will the force be if the distance between the asteroids is do

Alina [70]

To solve this problem, we use the formula:

F = G m1 m2 / r^2

where F is gravitational force, G is constant, m1 and m2 are masses while r is the distance between the two asteroids

Since G m1 m2 is constant, therefore:

F1 r1^2 = F2 r2^2

So if r2 = 2 r1:

(1,000,000 N) (r1^2) = F2 * (2 r1)^2

F2 = 250,000 N

It was divided by 4

3 0

3 years ago

lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of

bonufazy [111]

Answer:

The intensity of laser 2 is 4 times of the intensity of laser 1.

Explanation:

The intensity in terms of electric field is given by :

$U=\dfrac{1}{2}\epsilon_o E^2$

E is electric field

It means, $U\propto E^2$

In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.

Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,

$\dfrac{U}{U'}=(\dfrac{E}{E'})^2$

We have,

E'=2E

So,

$\dfrac{U}{U'}=\dfrac{E^2}{(2E)^2}\\\\\dfrac{U}{U'}=\dfrac{E^2}{4E^2}\\\\\dfrac{U}{U'}=\dfrac{1}{4}\\\\U'=4\times U$

So, the intensity of laser 2 is 4 times of the intensity of laser 1.

6 0

3 years ago

Plz help will give brainliest and ten points

Svetradugi [14.3K]

Hello yes whats the problem

3 0

3 years ago

What type of fingerprints is invisible to the naked eye?

luda_lava [24]

Well latent fingerprints are made of oil and sweat and generally materials that you can't see very easily, so it should be that.

Hope this helps :D

3 0

4 years ago

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago