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Brut [27]
4 years ago
7

Does the moon fall partly into earth's shadow when it is n "full"?

Physics
1 answer:
murzikaleks [220]4 years ago
5 0
Yes, <span> the moon fall partly into earth's shadow when it is in its full size</span>
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A 4kg watermelon is dropped from a height of 45m. What is the velocity of the watermelon just before it hits the ground?
Makovka662 [10]

Answer:

v=30 m/s

Explanation:

h - height

g - acceleration due to gravity=10

t - time

v- velocity

h =  \frac{1}{2}  \times g \times t {}^{2}

45 = 5t²

t² = 9

t=3 seconds

v=g×t

v=10×3

v=30 m/s

3 0
3 years ago
7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr
Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m

d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m

From the continuity equations in pipes we have to

A_1V_1 = A_2 V_2

Where,

A_{1,2} = Cross sectional Area at each section

V_{1,2} = Flow Velocity at each section

Then replacing we have,

(\pi r_1^2) v_1 = (\pi r_2^2) v_2

(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2

v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1

From Bernoulli equation we have that the change in the pressure is

\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)

7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)

7300= 8919.01 v_1^2

v_1 = 0.9m/s

Therefore the speed of flow in the first tube is 0.9m/s

6 0
4 years ago
A sphere of radius 6cm is moulded into a thin cylindrical wire of length 32cm. Calculate the radius of the
suter [353]

Answer:

4/3 pi R^3 = pi r^2 L     equating volume of sphere and wire

r = (4 R^3 / 3 * L)^1/2      solving for radius of wire

r = (4 * 6^3 / 3 * 32)^1/2

r = 9^1/2 cm = 3 cm = .03 meters

5 0
2 years ago
Express 9.39 x 109 seconds in terms of days.
Taya2010 [7]
108 681(nearest whole day)
7 0
4 years ago
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu
Rus_ich [418]

Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

The intensity is I  = 3.535 *10^{-6} \  W/m^2

Explanation:

From the question we are told that

    The temperature is  T = 2.81 \ K

Now  According to Stefan's law

        Power(P) =  \sigma  *  A  * T^4

Where  \sigma is the Stefan Boltzmann constant with value  \sigma  =  5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        I  =  \frac{P}{A}

=>      I  =  \frac{\sigma *  A  * T^4}{A}

=>      I  =  \sigma  *  T^4

substituting values

      I  = 5.67 *10^{-8}  *  (2.81)^4

       I  = 3.535 *10^{-6} \  W/m^2

       

4 0
3 years ago
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