T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !
Answer:
The acceleration of the ball is 666.67 m/s²
Explanation:
It is given that,
Mass of the baseball, m = 0.15 kg
Applied force to it, F = 100 N
We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :
F = ma



So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.
Answer:
F = M2 ω^2 R centripetal force of sun on planet
ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period
P = 2 pi (M2 * R / F)^1/2
F = G M1 M2 / R^2 gravitational force on planet
P = 2 pi {R^3 / (G M1)]^1/2
P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2
P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec
1 yr = 3600 * 24 * 365 = 3.15E7 sec
P = 3.9 / 3.2 = 1.2 years
Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
(a) 
The relationship between frequency and wavelength of an electromagnetic wave is given by

where
is the speed of light
is the frequency
is the wavelength
In this problem, we are considering light with wavelength of

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

(b) 
The period of a wave is equal to the reciprocal of the frequency:

And using
as we found in the previous part, we can find the period of this wave:
