Answer:
W = 8.01 × 10^(-17) [J]
Explanation:
To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:
W = q*V
where:
q = charge = 1,602 × 10^(-19) [C]
V = voltage = 500 [V]
W = work [J]
W = 1,602 × 10^(-19) * 500
W = 8.01 × 10^(-17) [J]
Liquids do not have a definite size and always take the shape of the container they're in.
The wavelength of a wave (λ) is given by λ

where c is the wave speed and f is the frequency
Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:
A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
ΔP=20 kg.m/s
Explanation:
Given data
Mass m=0.2 kg
Initial speed Vi=-44.5m/s
Final speed Vf=55.5 m/s
Required
Change in momentum ΔP
Solution
First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

Now we need to find the initial momentum
So

Substitute the given values

Now for final momentum

So the change in momentum is given as:
ΔP=P₂-P₁
![=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s](https://tex.z-dn.net/?f=%3D%5B%2811.1kg.m%2Fs%29-%28-8.9kg.m%2Fs%29%5D%5C%5C%3D20kg.m%2Fs)
ΔP=20 kg.m/s
Answer:
Explanation:
ignore air resistance
Let t be the time of fall for the dropped stone.
½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²
4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)
4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716
0 = 21.56t - 71.148
t = 71.148/21.56 = 3.3 s
h = ½(9.8)3.3² = 53.361 = 53 m
or
h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m