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Luden [163]
3 years ago
15

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Earth's pull?
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

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A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con
Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

T_B = 2T_A

As T = 2\pi\sqrt{\frac{m}{k}}, and the 2 springs have the same mass

2\pi\sqrt{\frac{m}{k_B}} = 2\pi\sqrt{\frac{m}{k_A}}

\sqrt{k_A} = 2\sqrt{B}

k_A = 4k_B

k_B = k_A/4

So A.The spring constant for B is one quarter of the spring constant for A. is the correct answer.

3 0
4 years ago
Compare and contrast series and parallel circuits?
dalvyx [7]

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

Hope It Helps!

6 0
3 years ago
HELP IS APPRECIATED!!
kobusy [5.1K]
Yeah i believe it is going to be c
8 0
4 years ago
The frequency of oscillation of a certain LC circuit is 299 kHz. At time t = 0, plate A of the capacitor has maximum positive ch
AnnZ [28]

Answer:

Explanation:

Given an LC circuit

Frequency of oscillation

f = 299 kHz = 299,000 Hz

AT t = 0 , the plate A has maximum positive charge

A. At t > 0, the plate again positive charge, the required time is

t =

t = 1 / f

t = 1 / 299,000

t = 0.00000334448 seconds

t = 3.34 × 10^-6 seconds

t = 3.34 μs

it will be maximum after integral cycle t' = 3.34•n μs

Where n = 1,2,3,4....

B. After every odd multiples of n, other plate will be maximum positive charge, at time equals

t" = ½(2n—1)•t

t'' = ½(2n—1) 3.34 μs

t" = (2n —1) 1.67 μs

where n = 1,2,3...

C. After every half of t,inductor have maximum magnetic field at time

t'' = ½ × t'

t''' = ½(2n—1) 1.67μs

t"' = (2n —1) 0.836 μs

where n = 1,2,3...

6 0
3 years ago
The electric potential at a point in space is -890 V. If a 0.0285 C charge is placed there, what will its potential energy U be?
BabaBlast [244]

Answer: -25.4

Explanation:

Acellus don’t forget the negative sign

5 0
3 years ago
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