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Ad libitum [116K]
2 years ago
11

How is the weight of an object in a spaceship near the moon related to the distance that the spaceship is from the moon?

Physics
1 answer:
Alex2 years ago
3 0
<span>As the distance between two objects
increases, the gravitational force decreases.
As the distance between two objects
</span><span>decreases, the gravitational force increases.</span> 271 N on the moon
i hope it works 
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13-16 that is where they’re located at
8 0
3 years ago
In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?
kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0
3 years ago
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In what way are mercury and venus similar ?
Neporo4naja [7]

Answer:

they are both planets they are both made of rock and  metel

Explanation:

3 0
3 years ago
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What is the volume of the cone?
dezoksy [38]

Answer:

42.417 cm³

Explanation:

The formula to find the volume of a cone is :

V = \frac{1}{3} × π r² h

Here,

r ⇒ radius ⇒ 3 cm

h ⇒ height ⇒ 4.5 cm

<u>Let us find it now.</u>

V = \frac{1}{3} × π r² h

V = \frac{1}{3} × π × 3 × 3 × 4.5

V = \frac{1}{3} × π × 9 × 4.5

V = \frac{1}{3} × π × 9 × 4.5

V = \frac{1}{3} × π × 40.5

V = \frac{1}{3} × 3.142 × 40.5

V = \frac{1}{3} × 127.251

V = <u>42.417 cm³</u>

4 0
2 years ago
A stone is tossed horizontally from the highest point of a 95 m building and lands 105 m from the base of the building. Ignore a
aalyn [17]

Answer:

A) t = 4.40 s , B)   v = 23.86 m / s ,  c)  v_y = - 43.12 m / s , D)  v = 49.28 m/s

Explanation:

This is a projectile throwing exercise,

A) To know the time of the stone in the air, let's find the time it takes to reach the floor

          y = y₀ + v_{oy} t - ½ g t²

as the stone is thrown horizontally  v_{oy} = 0

          y = y₀ - ½ g t²

          0 = y₀ - ½ g t²

          t = √ (2 y₀ / g)

          t = √ (2 95 / 9.8)

          t = 4.40 s

B) what is the horizontal velocity of the body

          v = x / t

          v = 105 / 4.40

          v = 23.86 m / s

C) The vertical speed when it touches the ground

          v_y = v_{oy} - g t

          v_y = 0 - 9.8 4.40

          v_y = - 43.12 m / s

the negative sign indicates that the speed is down

D) total velocity just hitting the ground

          v = vₓ i ^ + v_y j ^

          v = 23.86 i ^ - 43.12 j ^

Let's use Pythagoras' theorem to find the modulus

          v = √ (vₓ² + v_y²)

          v = √ (23.86² + 43.12²)

           v = 49.28 m / s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ (-43.12 / 23.86)

            θ = -61

7 0
3 years ago
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