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3 years ago

If time is tripled and work remains the same the power will.

Physics

1 answer:

VMariaS [17]3 years ago

7 0

Answer:

power will three times decrease (one-third)

Explanation:

P = W / t

thus when work is constant, power is inversely proportional to time

so, when time is tripled, power is decreased to one-third

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Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).

Oliga [24]

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

$\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}$

Where

P = Pressure at each point

r = Radius

$\eta =$ Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

From the problem two terms are given

$R_A = \frac{R_B}{2}$

$L_A = 2L_B$

Replacing we have to

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}$

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

6 0

3 years ago

A women weighs 857 N. What is her mass?

klemol [59]

Answer:

women based on height usually should way about 200 pounds which included muscle.

Explanation:

7 0

3 years ago

Read 2 more answers

In acceleration time graph curve upward shows _______ acceleration

kondor19780726 [428]

Answer:

rapid

Explanation:

it seems like it.

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2 years ago

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A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is

alexgriva [62]

Answer:

The speed of the flashlight at that point is 3.7 m/s

Explanation:

When an object of mass M is at a height H above the ground, the potential energy of the object is:

U = M*H*g

Where g is the gravitational acceleration, g = 9.8 m/s^2

And for an object with velocity v, the kinetic energy is:

K = (M/2)*v^2

We know that when the flashlight of mass 1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:

M = 1kg

H = 0.7m

g = 9.8 m/s^2

Replacing these in the equations, we get:

U = K

(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2

As the mass factor appears in both sides, we can remove it:

(0.7 m)*(9.8 m/s^2) = (v^2)/2

Now we can multiply both sides by 2:

2*(0.7 m)*(9.8 m/s^2) = v^2

Now let's apply the square root to both sides:

√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s

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2 years ago

In a real isothermal expansion, the temperature of the surroundings must be________the temperature of the gas.

Lady_Fox [76]

Must be isolated from the temperature of the gas

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2 years ago

If time is tripled and work remains the same the power will.​

If time is tripled and work remains the same the power will.