Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.
<span>The formula for force of attraction between any two bodies in the universe
F = GMm / r^2. (Newton's Universal law of Gravitation).
G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22 kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
= 3.8 x 10^8 m.
(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).
So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
Punch it all up in your calculator.
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.
F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!</span>
The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a
x₂ = 0.5*a*t² = 0.5*v°²/a
The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7
You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀.
Answer:
The rod`s charge must be positive, because the gravity force is pointing downwards and the electrostatic force must be pointing upwards (in order to balance the gravity force)
The charge is q_2 = 1.667 times 10^(-7) C
Explanation:
F_e = F_g
where F_g = m g and F_e= (1/4 pi e_0)*(q_1*q_2)/d^2,
please see the file attached for more details.
Option B is correct. If an incident light ray hits a flat, smooth object at 28°. It will reflect off at an angle of 28°.
<h3>What is the law of reflection?</h3>
The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.
The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.
When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.
If an incident light ray hits a flat, smooth object at 28 degrees, it will reflect off at an angle of 28°.
Hence, option B is correct
To learn more about the law of reflection, refer to the link;
brainly.com/question/12029226
#SPJ1
Answer:
50.2m/s
Explanation:
Using first equation of motion we have
V= u+gt
V= final speed
U= initial speed
So v = 1.2m/s + 9.8(5s)
50.2m/s
