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3 years ago

Please help answer the question in the screenshot. it's multiple choice..

Physics

1 answer:

Talja [164]3 years ago

7 0

Answer:

Since F = G * m1 * m2 / r^2

F = 6.67 * 19E-11 * 2.79 * 9.47 * 10E23 / (1.2 * 10^7)^2

F = 126 * 10E-2 N = 1.22 N

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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

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3 years ago

Phase, color, and ductility are all examples of what type of property?

Mama L [17]

Physical properties.

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3 years ago

Why do astronomers find it difficult to detect individual exoplanets?

Maurinko [17]

Light from the stars, because the orbits make it difficult to see them.

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2 years ago

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22

alexgriva [62]

1) The law of motion of the projectile is
$h(t) = 2+22.5 t-4.9 t^2$
To find the velocity, we should compute the derivative of h(t):
$v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t$
So now we can calculate the speed at t=2 s and t=4 s:
$v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s$
$v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s$
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
$v(t)=0$
So we have
$22.5-9.8 t=0$
And solving this we find
$t=2.30 s$

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
$h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m$

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
$2+22.5t -4.9 t^2 = 0$
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
$t=4.68 s$
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
$v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36 m/s$
with negative sign, because it is directed downward.

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3 years ago

A monochromatic light beam with a quantum energy value of 3.0 ev is incident upon a photocell. the work function of the photocel

otez555 [7]

1.4 ev I'd explain but I gotta roll!

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2 years ago

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