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loris [4]
3 years ago
14

The picture below was taken in the Maldives.

Physics
1 answer:
horrorfan [7]3 years ago
4 0

Answer:

it is c

Explanation:

a.p.e.x

i took the test

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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth
mylen [45]

Answer: 3.7 \times 10^{-4} N

Explanation:

The gravitational pull between two object is given by:

F = G\frac{Mm}{r^2}

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N

4 0
3 years ago
Read 2 more answers
A 500 g ball swings in a vertical circle at the end of a 1.4-m-long string. when the ball is at the bottom of the circle, the te
sergij07 [2.7K]

A 500 g ball swings in a vertical circle at the end of a 1.4-m-long string. when the ball is at the bottom of the circle, the tension in the string is 18 n.

6 0
3 years ago
A trampoline has a spring constant of 3430 N/m how far will the trampoline sink when a 70 kg person steps on it
Dominik [7]
F = ma = -kx

a = 9.81 m/s²
k = 3430 N/m
m = 70 kg

x = - ma/ k = 0.2m
3 0
3 years ago
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A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).
yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

f = n\frac{v}{2L}

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

f = n\frac{v}{2L}\\L = \frac{nv}{2f}

The radius between the two frequencies would be 4 to 5,

\frac{528Hz}{660Hz}= \frac{4}{5}

4:5

Therefore the frequencies are in the ratio of natural numbers.  That is

4f = 528\\f = \frac{528}{4}\\f = 132Hz

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m

Therefore the length of the pipe in the second harmonic is 2.6m

7 0
3 years ago
Which option is an example of reflection?
katovenus [111]

Answer:

A

Explanation:

sound waves hit the canyon walls, then bounce back towards you. That's why you hear an echo

6 0
2 years ago
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