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4 years ago

8

Radio telescopes have poorer angular resolution than optical telescopes because radio waves have a much longer wavelength than o

ptical waves. Group of answer choices True False

1 answer:

Kitty [74]4 years ago

8 0

Answer:

True

Explanation:

Radio waves have a much longer wavelength than optical waves

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In order to be considered scientific, evidence must be:

zhenek [66]

Reviewed by many sources. for example, it must be reviewed by scientist too

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4 years ago

I need help on 2/3 please asap thx ♥️

ch4aika [34]

<u>questions 2</u>

F=m

therefore

a=F/m

a=408/68

=6m/s^2.

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4 years ago

Shakina and Juliette set the car's initial velocity to zero and set the acceleration to +1.2 m/s2, then clicked "start." Answer

Katarina [22]

Let's start by differentiating the terms distance and displacement. They both refer to the length of paths. Distance only accounts for the total length regardless of the path taken. Displacement measures the linear path from the starting point to the end point. So, it does not necessarily follow the actual path. However, for this problem, assuming that the path is just in one direction, displacement and distance would just be equal. The equation would be:

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3 years ago

What is the magnitude of speed of a person goes 75 meters east followed by 20 meters west in 4 seconds

marshall27 [118]

Answer:23.75 m/sec

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3 years ago

Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S an

insens350 [35]

Answer : The correct option is, (d) 4 times

Solution :

According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.

Formula used :

$F=k_e\frac{q_1q_2}{r^2}$

where,

F = electrostatic force of attraction or repulsion

$k_e$ = Coulomb's constant

$q_1$ and $q_2$ are the charges

r = distance between two charges

First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'

$F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2$ ..........(1)

Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.

$F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}$ ...........(2)

Equation equation 1 and 2, we get

$\frac{F_{sq}}{F_{sp}}=\frac{1}{4}$

$F_{sq}=4\times F_{sp}$

Therefore, the force exerted between S and q is 4 times the force exerted between S and p.

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3 years ago

Read 2 more answers