<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>
Answer:
Mass and height
Explanation:
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about ![9.8 m/s2.](https://tex.z-dn.net/?f=9.8%20m%2Fs2.)
Which is represented as;
![PE_g=mgh](https://tex.z-dn.net/?f=PE_g%3Dmgh)
stands for gravitational potantial energy,
m stands for mass of object,
g is the gravitational constant and
h is the height.
Here we see that mass of object and height is directly proportional to the gravitational potential energy.
That means increasing in mass and height will result in increasing gravitational potential energy.
Answer:
b) d = 0.71 Km
Explanation:
Car kinematics
Car 1 moves with uniformly accelerated movement
Formula (1)
d: displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Equivalences:
1mile = 1609.34 meters
1 hour = 3600s
1km = 1000m
Known data
![v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}](https://tex.z-dn.net/?f=v_o%20%3D%2069%5Cfrac%7Bmile%7D%7Bhour%7D%20%2A1609.34%5Cfrac%7Bmeter%7D%7Bmile%7D%20%2A%5Cfrac%7B1%7D%7B3600%7D%5Cfrac%7Bhour%7D%7Bs%7D%3D30.8%20%5Cfrac%7Bm%7D%7Bs%7D)
![v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}](https://tex.z-dn.net/?f=v_f%3D%20%5Cfrac%7B69%7D%7B2%7D%20%5Cfrac%7Bmile%7D%7Bhour%7D%20%3D34.5%5Cfrac%7Bmile%7D%7Bhour%7D%3D15.4%20%5Cfrac%7Bm%7D%7Bs%7D)
a = -0.5 m/s²
Distance calculation
We replace data in the Formula (1)
![15.4^2 = 30.8^2+2(-0.5)*d](https://tex.z-dn.net/?f=15.4%5E2%20%3D%2030.8%5E2%2B2%28-0.5%29%2Ad)
![2(0.5)*d = 30.8^2 - 15.4^2](https://tex.z-dn.net/?f=2%280.5%29%2Ad%20%3D%2030.8%5E2%20-%2015.4%5E2)
![d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m](https://tex.z-dn.net/?f=d%20%3D%5Cfrac%7B%2030.8%5E2%20-%2015.4%5E2%7D%7B2%280.5%29%20%7D%3D%20717.6m)
![d = 717.6 m\frac{1km}{1000m} = 0.7176Km](https://tex.z-dn.net/?f=d%20%3D%20717.6%20m%5Cfrac%7B1km%7D%7B1000m%7D%20%3D%200.7176Km)
Answer:
I believe its A: Sports biomechanics.