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Bess [88]
2 years ago
13

A concave mirror creates a real, inverted image 16.0 cm from its surface. If the image is 3.5 times larger, how far away is the

object? Use the GUESS method to solve the problem and show all your work. Record you final answer using the correct number of significant digits
Physics
1 answer:
Juliette [100K]2 years ago
5 0

Answer:

3.6, make sure you study

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Mars has twice the mass of Mercury and is 4 times further away from the Sun. Calculate theratio of the gravitational force from
svetoff [14.1K]

Answer:

F(Mars) = 2 G m M / (4 R)^2   force of Sun on Mars

F(Merc) = G m M / R^2    force of force of Sun on Mercury

R = distance of Sun from Mercury, m = mass of Mercury

F(Merc) / F(Mars) = 4^2 / 2 = 8

6 0
2 years ago
A ball bounces on the ground. How do the ball and the ground act on each other?(1 point)
liubo4ka [24]

Answer: A is your best answer.

Explanation:

It should be A because the when the ball bounces on the ground the ground will give it force to bounce again but also it wont go as high as it first did. Hope this helps:))

3 0
3 years ago
Read 2 more answers
A 24-V battery is powering a light bulb with a resistance of 3.0 ohms. What is the current flowing through the bulb? A) 7.20 A B
Usimov [2.4K]

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=24/3=8 A

The correct answer is  B


3 0
3 years ago
Assume that a gravitational anomaly in the solar system has shifted a field of asteroids into Earth’s orbit, and the field is no
Mandarinka [93]

Answer:

An asteroid is a minor planet of the inner Solar System. Historically, these terms have been applied to any astronomical object orbiting the Sun.

7 0
2 years ago
The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?
RoseWind [281]

Answer:

E = 2.9775\times10^5 J

Explanation:

Given:  The latent heat of fusion for Aluminum is L = 3.97\times10^5  J/Kg

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J

Therefore, energy required to melt 0.75 Kg aluminum

E = 2.9775\times10^5 J

5 0
3 years ago
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