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VLD [36.1K]
2 years ago
8

Q C During a solar eclipse, the Moon, the Earth, and the Sun all lie on the same line, with the Moon between the Earth and the S

un. (c) What force is exerted by the Sun on the Earth?
Physics
1 answer:
lianna [129]2 years ago
3 0

The force exerted by the Sun on the Earth is =3.557 \times 10^{22}

<h3>What is the Netwon law of the gravity state?</h3>

Newton's law of gravity states that the force executed by a body on another body is calculated

$F=\frac{G \cdot m_1 \cdot m_2}{d^2}$

<h3>What is universal gravitation?</h3>

The universal gravitation constant

$G=6 \cdot 67 \times 10^{-11} \cdot \mathrm{m}^3 \cdot 8^{-2} \cdot \mathrm{kg}^{-1}$

Here we know that

The mass of the sun is $m_s=2.0 \times 10^{30} \mathrm{~kg}$ is the

The mass of the moon is $m_M=7.3 \times 10^{22} \mathrm{~kg}$

The mass of the Earth is $m_E=6.0 \times 10^{24} \mathrm{kg}$

The distance between the Sun and the Earth is

$$d_{S E}=1.5 \times 10^{11} \mathrm{~m}$$

The distance between the earth and the moon is

$$d_{E M}=3.85 \times 10^8 \mathrm{~m} \text {. }$$

Here the calculation is:

F_{S E}=\frac{G \cdot m_S \cdot m_E}{\left(d_{S E}\right)^2}

$=\frac{6.67 \times 10^{-11} \times 2.0 \times 10^{30} \times 6.0 \times 10^{24}}{\left(1.5 \times 10^{11}\right)^2}$

$=\frac{80.04 \times 10^{43}}{1.5^2 \times 10^{22}}=\frac{80.04 \times 10^{21}}{2.25}$

=3.557 \times 10^{22}

To learn more about quadratic equations, refer: brainly.com/question/21570007

#SPJ4

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) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

f= 466 Hz

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s

The angular frequency instead is given by

\omega = 2\pi f

And substituting

f = 466 Hz

We find

\omega = 2\pi (466 Hz)=2926.5 rad/s

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz

While the angular frequency is given by:

\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s

(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

7 0
3 years ago
I need help please and thanks
asambeis [7]

Answer:

c

Explanation:

6 0
2 years ago
Name 3 different types of graphs that can be used to plot data.
siniylev [52]

Answer:

Bar graph

Pie graph

Line Graph

Explanation:

mark as brainliest and drop some thanks!!!

7 0
3 years ago
Read 2 more answers
How much time does it take a dropped object to fall 180 m on Earth?
rusak2 [61]

Answer:

6s

Explanation:

Assume it is dropped from rest and the gravitational acceleration is 10

By the equation of motion under constant acceleration:

s=ut+\frac{1}{2} at^2

180 = (0)t+10(t^2)/2

t = 6 or -6 (rejected)

t = 6 s

7 0
3 years ago
An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17
Softa [21]

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

3 0
3 years ago
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