When I went through with the math, the answer I came upon was:
<span>6.67 X 10^14 </span>
<span>Here is how I did it: First of all we need to know the equation. </span>
<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
<span>(3.0 X 10^8 m/s) = (frequency)(450nm) </span>
<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>
<span>Divide the speed of light by the wavelength. </span>
<span>(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- </span>
<span>Answer: 6.67 X 10^14 s- hope this helps</span>
Answer:
B) The amount of time the torque is applied to the disk, because the time interval is related to the angular impulse of the disk.
Explanation:
Angular impulse = Torque x time
= change in angular momentum
So,
Torque x time = change in angular momentum
change in angular momentum = Torque x time
Torque is already known .
Hence to know the change in angular momentum what is needed to know is time duration of torque acting on the body .
Answer: 172
Explanation: Because yeah
<h3><u>Answer;</u></h3>
just before it reaches the ground
<h3><u>Explanation;</u></h3>
- Kinetic energy is the energy possessed by a body or an object in motion.
- <em><u>Kinetic energy is given by 1/2mv², where m is the mass of the object and V is the velocity of the body. Thus, kinetic energy depends on the velocity of the body if mass is kept constant.</u></em>
- <em><u>As soon as the ball leaves the racket it has more kinetic energy and zero potential energy. As it moves up its velocity decreases, and thus the kinetic energy is being converted to kinetic energy up to maximum height reached where kinetic energy will be zero since the velocity is zero.</u></em>
- <em><u>When the ball is going down the potential energy will be converted to kinetic energy up to a point just before it hits the ground, where kinetic energy is maximum since the velocity of the ball is maximum, due to gravitational acceleration.</u></em>