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3 years ago

9

The gravity of Neptune is about 1.1 times the gravity of earth. How will the mass of an object on Neptune compare with its mass

on earth

2 answers:

Roman55 [17]3 years ago

7 0

I think the gravity doesn't affect the mass of an object. Only it's weight can be compared

frosja888 [35]3 years ago

4 0

Gravity has no effect on mass.

But the same mass will WEIGH 1.1 times as much on Neptune if you take it there.

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An internal resistance of 5 ohm and a battery of 15 ohm is connected to a resistance of 20 ohm calculate the electric current

Allushta [10]

Answer:

.6 A

Explanation:

Battery 15 VOLTS

V = IR

V / R = I

15 / ( 5+20) = .6 amps

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2 years ago

You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that

loris [4]

Questions Diagram is attached below

Answer:

$T=2.08s$

Explanation:

From the question we are told that:

Speed of Train $V=3.0m.s$

Angle $\theta=12\textdegree$

Height of window $h_w=0.90m$

Width of window $w_w=2.0m$

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$b=4.23$

Therefore

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$d_a=2.0+4.23$

$d_a=6.23$

Therefore

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5 0

2 years ago

Will an object with more mass roll faster down a hill?

inna [77]

Over time, yes. It will over time gain more momentum

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3 years ago

How do we know gravity is a field force?

Papessa [141]

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Explanation:

7 0

3 years ago

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun = 2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0

3 years ago