Answer:
53.33 seconds
Explanation:
From the question;
- Power of the motor is 75 kW or 75000 W
- Depth or height is 150 m
- Volume of water is 400 m³
We are required to determine taken to raise the water from the given height.
We know that density of water is 1000 kg/m³
Therefore;
Mass of water = 400 m³ × 1000 kg/m³
= 4.0 × 10^5 kg
Thus, force required to raise the water;
= 4.0 × 10^5 kg × 10 N/kg
= 4.0 × 10^6 N
To determine the time;
we use the formula;
Time = work done ÷ power
= (4.0 × 10^6 N × 150 m) ÷ 75000 Joules/s
= 53.33 seconds
Therefore, time taken to raise the water is 53.33 seconds
Answer: 9.0 atm
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
where,
are initial pressure and volume.
are final pressure and volume.
We are given:
Putting values in above equation, we get:
Thus new pressure of 150 ml of a gas that is compressed to 50 ml is 9.0 atm
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
Increasing the pressure of gas is like exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume.
Hope this helps!
