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3 years ago

If you were a lunar inhabitant what kinds of eclipses might you expect to see

1 answer:

5 0

Hello
You expect to see:

1) A solar eclipse, when the earth is between the sun and the moon: the shadow of the earth is projected on the moon surface, and the sun will appear dark

2) An "earth eclipse", when the moon is between the sun and the earth: the shadow of the moon is projected on the earth surface, therefore the earth will appear dark.

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A light bulb has a power output of 80 W and it has been left on for 90 seconds. How much work was done?

Karo-lina-s [1.5K]

7200 joules of heat and light energy was dissipated into the air. But no work was done ... no force moved through no distance.

3 0

3 years ago

A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin

erastovalidia [21]

Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫ dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5 / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

8 0

3 years ago

On Venus, the acceleration due to gravity is 8.87 m / s 2 . How far would a 17 g rock fall from rest in 6.5 s if the only force

algol [13]

Answer:

187.38 m

Explanation:

Using the equation of motion

s = ut + 1/2gt²...................... Equation 1

Where s = distance of fall, u = initial velocity of the rock, t = time taken for the rock to fall from rest, g = acceleration due to gravity of venus.

Given: u = 0 m/s ( from rest), t = 6.5 s, g = 8.87 m/s².

substituting into equation 1

s = 0(6.5) + 1/2(8.87)(6.5)²

s = 0 + 374.7575/2

s = 187.38 m.

Hence the rock will fall 187.38 m

7 0

4 years ago

Two objects of mass m move in opposite directions toward each other. The green object moves at velocity v, and the blue object m

NeX [460]

. The velocity of a mass attached to a spring is given by v = (1.5 cm/s) sin(ωt + π/2), ..... Which of the following is the motion of objects moving in two dimensions

5 0

4 years ago

Read 2 more answers

A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit

Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0

3 years ago