I see the word "when..." kind of fading out at the end of the first line.
Whatever comes after it may be important.
If you're just supposed to copy the expression into the box,
then the problem is that you left the 'e' out of it.
I'm guessing that you're supposed to enter whatever the expression becomes
when either N₀ or ' t ' has some special value that's in the first line.
Just taking a wild guess here . . . . .
If it's "Enter the expression ..... , when t=0 ." ,
then the correct answer in the box is N₀ .
But that's just a wild guess. As I pointed out, you cut off
the picture in the middle of the word 'when', and I've got
a hunch that there's something important after it.
Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done
The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.
The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy
The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.
Answer:
525 kg.m/s
Explanation:
★ Momentum = Mass× Velocity
→ P = (7.5 × 70) kg.m/s
→ P = (75 × 7) kg.m/s
→ <u>P</u><u> </u><u>=</u><u> </u><u>5</u><u>2</u><u>5</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>
