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Sav [38]
2 years ago
5

The space shuttle approaches earth at a speed of 2.89*10^4 km/h when it sends a radio signal to earth 1.00*10^5 km away. The rad

io signal will reach earth in ____ seconds
Physics
1 answer:
zysi [14]2 years ago
6 0

Answer:

3 secods later

Explanation:

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If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil
Hoochie [10]

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second  and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length  of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>

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2 years ago
Based on this passage, what is campylobacter?
alexdok [17]
I believe the correct answer from the choices listed above is option 4. Base from the passage given above, it is very clear that the Campylobacter is a bacteria used to test in <span>identifying drug-resistant bacteria. Hope this answers the question. Have a nice day.</span>
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3 years ago
Read 2 more answers
Two boys wearing roller skates are standing on a smooth surface with the palms of their hands touching and their arms bent, as s
vlada-n [284]

Boy X and Boy Y both move backward in opposite directions.
8 0
2 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
2 years ago
A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S
Black_prince [1.1K]

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

8 0
2 years ago
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