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notsponge [240]
3 years ago
13

What happens to the length of the day due to refraction of light?​

Physics
1 answer:
professor190 [17]3 years ago
8 0

One of the consequences of this effect of refraction is that contrary to what most people think, the day and night is not exactly 12 hours long on the days of the Equinoxes. Because of refraction of sunlight, daytime is just slightly longer than night time on March Equinox and September Equinox. This effect also has an impact on the time that the Moon can be seen before Moonrise and after Moonset.

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A 10 kg block is attached to a light cord that is wrapped around the pulley of an electric motor, as shown above. If the motor r
Anika [276]

Answer:

156.8 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 10 kg

Height (h) = 8 m

Time (t) = 5 s

Power (P) =?

Next, we shall determine the energy used by the motor to raise the block. This can be obtained as follow:

Mass (m) = 10 kg

Height (h) = 8 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 10 × 9. 8 × 8

E = 784 J

Finally, we shall determine the power output of the motor. This can be obtained as illustrated below:

Time (t) = 5 s

Energy (E) = 784 J

Power (P) =?

P = E/t

P = 784 / 5

P = 156.8 Watts

Therefore, the power output of the motor is 156.8 Watts

7 0
2 years ago
To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

3 0
2 years ago
An archer tests various arrowheads by shooting arrows at a pumpkin that is suspended from a tree branch by a rope, as shown to t
erik [133]

Answer:

Bounce 1 ,  pass 3,   emb2

Explanation:

(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle.  So it is  Bounce1, pass3, emb2.  

6 0
3 years ago
Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomical o
inysia [295]

Answer:

Telescope

Explanation:

Telescope is usually defined as an optical instrument that is commonly used to observe the objects in a magnified way that are located at a large distance from earth. These telescopes are comprised of lenses and curved mirrors that are needed to be arranged in a proper way in order to have a prominent look. It is commonly used by the astronomers.

This was first constructed by Hans Lippershey in the year 1608.

6 0
2 years ago
Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from a
riadik2000 [5.3K]

Answer:

Check the explanation

Explanation:

The beat frequency is

df = f2 - f1

the wavelength is

lamda1 = (v/f1)

and lamda2 = (v/f2)

where v = 340 m/s,f1 = 25.0 kHz and f2 = 20.0 kHz

8 0
2 years ago
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