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2 years ago

Which planet is approximately 20 times farther from the sun than earth is answer\?

1 answer:

6 0

The answer is Uranus. Its farthest distance from the Sun (aphelion) as point of reference is approximately 3 billion kilometers. Whereas Earth's distance from the Sun is about 150 million kilometers. Dividing Earth's distance to sun with the Uranus' distance to the Sun will give us the quotient of 20. Hence, Uranus is the planet that is 20 times farther from the Sun than the Earth is.

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An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Lynna [10]

Answer:

$\boxed{\sf Magnification \ (m) = 2}$

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

$\boxed{\bold{\sf Magnification \: (m) = - \frac{Image \: distance \: (v)}{Object \: distance \: (u)}}}$

Substituting values of Image distance(v) & Object distance (u) in the equation:

$\sf \implies m = - \frac{( - 50)}{25}$

-(-50) = 50:

$\sf \implies m = \frac{50}{25}$

$\sf \implies m = \frac{2 \times \cancel{25}}{ \cancel{25}}$

$\sf \implies m = 2$

4 0

2 years ago

An astronaut hits a golf ball of mass m on the Moon, where there is no atmosphere and the acceleration due to gravity is g 6 , w

sasho [114]

Answer:

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Explanation:

Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v

As we know,

m*v = Integration of F.dt with limits 0 to T

Ft = mv

F = mv/t

The magnitude of the average force exerted by the club on the ball during contact = mv/t

5 0

2 years ago

A 2 kg stone moves with a speed of 1 m/s. A second 2 kg stone is moving twice as fast. Compare their kinetic energies.

alekssr [168]

D
is the answer
Well it should be

5 0

2 years ago

Read 2 more answers

1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

Juli2301 [7.4K]

The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

Time, t = 2s

Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

F = 6975N

Therefore, the force exerted on the car during this stop is 6975N

6 0

2 years ago

14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

Vladimir79 [104]

The relationship between resistance R and resistivity $\rho$ is
$R= \frac{\rho L}{A}$
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m$
and the cross section is
$A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2$

From the first equation, we can then find the length of the wire when $R=4.8 \Omega$ (copper resistivity: $\rho = 1.724 \cdot 10^{-8} \Omega m$ )
$L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m$

4 0

2 years ago