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erik [133]
3 years ago
15

The diameter of the moon is a little less than the distance across the United States. Please select the best answer from the cho

ices provided T F
Physics
2 answers:
dlinn [17]3 years ago
6 0

Answer:

true

Explanation:

The diameter of the Moon is 3474 km. The distance across the United States, from Florida to Washington, is 4509.382 km.

katrin2010 [14]3 years ago
5 0

It is true that the diameter of the moon is a little less than the distance across the United States.

Answer: Option A

<u>Explanation: </u>

Moon is the natural satellite of our Earth. It revolves around the Earth due to the gravitational force acting on the moon. The moon also has its own gravitational force. The gravitational force of moon is \frac{1}{6} t h of the gravitational force on Earth.

The mass of moon is 7.34 \times 10^{22} kg while the mass of the Earth is 5.972 \times 10^{24} kg. So the moon’s mass is very very less than that of Earth. So the diameter of moon is found as 3,472 km while the diameter of Earth is 12,742 km.

Thus the dimension of moon is \frac{1}{4} t h of the dimension of the Earth. Even the diameter of moon is little less than the distance across the United States as the distance across United States is 4509 km.

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) =  7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0
3 years ago
Name the fundamental units involved in pascal<br>​
gtnhenbr [62]

Explanation:

Hey there!

Here,

Pascal is a unit of pressure.

pressure =  \frac{f}{a}

Now, As per the formula the units are:

kg, m and s^2.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

5 0
3 years ago
An airplane was moving 1200 km per minute traveling in a north-to-south direction when they were spotted by air traffic control.
Feliz [49]

Its velocity was 20 km/s southward

7 0
3 years ago
Read 2 more answers
Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *
tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2

So, the required acceleration is 0.45 m/s².

4 0
3 years ago
A 2.0 kg mass moving at 5.0 m/s suddenly collides head-on with a 3.0 kg mass at rest. If the collision is perfectly inelastic, h
liberstina [14]

Answer:

correct answer is (c) 15 J

Explanation:

given data

mass m1 = 2 kg

velocity V1 = 5 m/s

mass other  = 3 kg

so mass m2 = 2+ 3 kg = 5 kg

solution

we will apply here conservation of momentum:

m1V1 = m2V2  ..........................1

put here value and we get  velocity v2

(2.0) × (5.0) = (2.0 + 3.0) × V

solve it we get

10 = 5 × V 2

V2 = 2.0 m/s

so here kinetic energy will be

KE = ½ × m × v²

so

∆KE = ½ × m1 × (v1)²  -  ½  × m2 × (v2) ²

∆KE = 0.5 × 2 × 25 - 0.5 × 5 × 4

 ∆KE = 25 - 10

∆KE  = 15 J

4 0
3 years ago
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