Answer:
Specific heat capacity means that in order to raise copper by 1C per kg, it needs 380J of heat energy.
Answer:
The answer to your question is:
Explanation:
There are two kinds of cell transport passive transportation and active transportation.
Passive transportation does not need energy because molecules move from higher concentration to lower concentration.
Active transportation needs energy because molecules moves against concentration.
a. facilitated diffusion It's an example of passive transportation so this answer is wrong.
b. passive transport Molecules move in favor of concentration so this answer is wrong.
c. osmosis is another example of passive transport so this answer is wrong.
d. simple diffusion it's another example of passive transport, so it's wrong this answer.
e. active transport this is the right answer.
Time required : 3 s
<h3>Further explanation
</h3>
Power is the work done/second.
![\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s](https://tex.z-dn.net/?f=%5Ctt%20P%3D%5Cdfrac%7BW%7D%7Bt%7D%5C%5C%5C%5CP%3Dpower%2Cj%2Fs%2Cwatt%5C%5C%5C%5CW%3Dwork%2C%20J%5C%5C%5C%5Ct%3Dtimes%3Ds)
To do 33 J of work with 11 W of power
P = 11 W
W = 33 J
![\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}](https://tex.z-dn.net/?f=%5Ctt%20t%3D%5Cdfrac%7BW%7D%7BP%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B33%7D%7B11%7D%5C%5C%5C%5Ct%3D%5Cboxed%7B%5Cbold%7B3~s%7D%7D)
Answer:
at the highest point of the path the acceleration of ball is same as acceleration due to gravity
Explanation:
At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest
So here we will say that at the highest point of the path the speed of the ball comes to zero
now by the force diagram we can say that net force on the ball due to gravity is given by
![F_g = mg](https://tex.z-dn.net/?f=F_g%20%3D%20mg)
now the acceleration of ball is given as
![a = \frac{F_g}{m}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_g%7D%7Bm%7D)
![a = \frac{mg}{m} = g](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bmg%7D%7Bm%7D%20%3D%20g)
so at the highest point of the path the acceleration of ball is same as acceleration due to gravity