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4 years ago

The turbines can be seen inside this hydroelectric dam. Why are they located at that particular height?

Physics

2 answers:

Yakvenalex [24]4 years ago

5 0

Answer:

Explanation:

the answer is number three

Mariulka [41]4 years ago

3 0

Answer: The answer is number is two

Explanation: Because as it says there is a large drop, so the water will still have a lot of kinetic energy to transfer to the turbine.

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Which of the following would likely be the best reflector of heat energy

mr Goodwill [35]

Ok so B would be the best surface to reflect heat energy as it's polished and does not disturb the wave of energy like the others would. Also in therms of colour it's better because it's a lighter colour than navy ( the darker colour of the spectrum ). This matters as darker colours absorb light where light colour reflect it.
Hope this helps :).

6 0

3 years ago

Read 2 more answers

Our atmosphere exerts a pressure of 14.7 lb/in2. What is the force on one square foot of the earth’s surface caused by the atmos

Studentka2010 [4]

Answer:

The force on one square foot of the earth’s surface caused by the atmosphere is 101.4 kN.

Explanation:

To force is given by:

$F = P*A$ (1)

Where:

P: is the pressure = 14.7 lb/in²

A: is the area = 1 m²

By entering the above values into equation (1) we have:

$F = P*A = 14.7 \frac{lb}{in^{2}}*\frac{4.45 N}{1 lb}*\frac{1550 in^{2}}{1 m^{2}}*1 m^{2} = 101.4 kN$

Therefore, the force on one square foot of the earth’s surface caused by the atmosphere is 101.4 kN.

I hope it helps you!

3 0

3 years ago

A 87 kgkg skydiver can be modeled as a rectangular "box" with dimensions 18 cmcm ×× 47 cmcm ×× 180 cmcm . If he falls feet first

garik1379 [7]

Answer:

Therefore the terminal velocity = 1.45 m/s

Explanation:

Terminal velocity: Terminal velocity is the highest velocity of an object when it falls from rest trough a media.

$V_t=\sqrt{\frac{2w}{c_d\rho A}}$

$V_t$ = terminal velocity

w = weight of the object = mg

$c_d$ = drag coefficient=0.80

A= frontal area

$\rho$ = media density = 1.2 kg/m³

m = mass = 8 kg

g= acceleration due to gravity = 9.8 m/s²

Front area = length× breadth

= (18×47)cm²

=846 cm²

Therefore the terminal velocity

$v_t= \sqrt{\frac{2\times 87\times 9.8}{0.80\times 846 \times 1.2}$

=1.45 m/s

Therefore the terminal velocity = 1.45 m/s

8 0

3 years ago

A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because

Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

$\tau = F*d$

Where

F= Force

d = Distance

Our values are given as,

$m_1 = 0.32Kg$

$m_2 = 0.4Kg$

$d = 8.7*10^{-2}m$

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

$\tau = T_2-T_1$

$\tau = F_2*d-F_1*d$

$\tau = m_2g*d-m_1*g*d$

$\tau = (m_2-m_1)g*d$

$\tau = (0.4-0.32)(9.8)(8.7*10^{-2})$

$\tau = 0.068N\cdot m$

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is $\tau = 0.068N\cdot m$

8 0

3 years ago

An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line

Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given by

$V=-2K\lambda ln\left (\frac{r'}{r} \right )$

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

$qV = 0.5m(u^{2}-v^{2})$

By substituting the values, we get

$V=\frac{mu^{2}}{2q}$

$-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}$

$- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }$

$- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }$

$- ln\left ( \frac{r'}{r} \right )=0.35$

$\frac{r'}{r} =e^{-0.35}$

$\frac{r'}{r} =0.7047$

r' = 14.5 x 0.7047 = 10.22 cm

5 0

3 years ago