Answer:
Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.
Explanation:
When a ball is thrown upward under the free fall action of gravity, it starts to loose its Kinetic Energy as it moves upward. As the ball moves in upward direction, its kinetic energy gradually converts into its potential energy. As a result the speed of the ball starts to decrease as it moves up. Therefore, at the highest point during its motion, the velocity of ball becomes zero and it stops at the highest point for a moment, and then it starts to fall back down, under the influence of gravitational force.
Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation <u>when the ball reaches the highest point of its motion.</u>
Answer:
It doesn’t really relate
Explanation:
heavier load the parachute must be moving faster to match the downward force of the greater load
and approx terminal velocity when the parachute is open
velocity for Ping pong ball with parachute = 9m/s
velocity for Soccer ball with parachute = 15m/s
velocity for Golf ball with parachute =24m/s
velocity for Watermelon with parachute = 25m/s
so weight of an object doesn’t really realted how fast it falls with a parachute
I'm not 100% sure that this is right, however I believe it is the firefly's thorax
Answer:
the answer to this question is A
Answer:
Assuming the emissivity is 1.0, the temperature of the wood stove will be about 417 C
Explanation:
You can use the Stefan-Boltzmann formula tying energy with temperature of a blackbody:
P - rate of radiation
is the blackbody's emissivity
is the Stefan-Boltzmann constant
A - area
T - temperature
So,
![P=\epsilon \sigma AT^4\implies T = \sqrt[4]{\frac{P}{\epsilon \sigma A}}\\T = \sqrt[4]{\frac{18000W}{1.0\cdot 5.67\cdot 10^{-8} \frac{W}{m^2\cdot K^4}\cdot 1.4 m^2}}\approx 690K \approx 417^\circ C](https://tex.z-dn.net/?f=P%3D%5Cepsilon%20%5Csigma%20AT%5E4%5Cimplies%20T%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7BP%7D%7B%5Cepsilon%20%5Csigma%20A%7D%7D%5C%5CT%20%3D%20%5Csqrt%5B4%5D%7B%5Cfrac%7B18000W%7D%7B1.0%5Ccdot%205.67%5Ccdot%2010%5E%7B-8%7D%20%5Cfrac%7BW%7D%7Bm%5E2%5Ccdot%20K%5E4%7D%5Ccdot%201.4%20m%5E2%7D%7D%5Capprox%20690K%20%5Capprox%20417%5E%5Ccirc%20C)
Assuming the emissivity ( a number between 0 and 1) is 1.0, the temperature of the wood stove will be about 417 C. The emissivity number was not given in your question. If you determine what it is, you can divide this result by the fourth power of that value to get an updated result.