The location of the strongest magnetic force is the

A pump is used to transport water to a higher reservoir. if the water temperature is 15ºc, determine the lowest pressure that ca

Reika [66]

Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called cavitation.

So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at $15^{\circ}$ is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.

7 0

3 years ago

The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r

maw [93]

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

6 0

3 years ago

Definition of distance travelled?

Mekhanik [1.2K]

Answer:

distance traveled is a total length of the path traveled between two positions.

6 0

3 years ago

Read 2 more answers

A gymnast dismounts off the uneven bars in a tuck position with a radius of 0.3m (assume she is a solid sphere) and an angular v

kifflom [539]

Here we will say that there is no external torque on the system so we will have

$L_i = L_f$

here we know that

$L_i = I_1\omega_1$

where we know that

$I_1 = \frac{2}{5}mr^2$

Also we know that

$I_2 = \frac{1}{12}mL^2$

initial angular speed will be

$\omega_1 = 2\pi(2rev/s) = 4\pi rad/s$

now from above equation

$\frac{2}{5}mr^2 (4\pi) = \frac{1}{12}mL^2 \omega$

$0.4(0.3)^2(4\pi) = \frac{1}{12}(1.5)^2\omega$

$0.452 = 0.1875 \omega$

now we have

$\omega = 2.41 rad/s$

so final speed will be 2.41 rad/s

6 0

3 years ago

Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.8 s after the eleva

Delicious77 [7]

Answer:

v= 4.0 m/s

Explanation:

When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
Gravity (which we call weight near the Earth's surface) can be calculated as follows:

$F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)$

According to Newton's 2nd Law, it must be met the following condition:

$F_{net} = F_{g} -F_{n} = m*a\\ F_{net} = 930 N - 830 N = 100 N = 95 Kg * a$

As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.

Solving for a, we get:

$a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} = 1.05 m/s2$

We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

$v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s$

Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
According to Newton's 2nd Law, if net force is 0, the object is either or at rest, or moving at a constant speed.
As the elevator was moving, the only choice is that it is moving at a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.

3 0

3 years ago