3 years ago

If jupiter were scaled to the size of a basketball, earth would be the closest to the size of

1 answer:

7 0

Jupiter's diameter is 11.2 times larger than Earth. This means that Earth would be relatively small. Based on searches around the internet from reliable sources, I can conclude that the earth would be the closest to the size of a marble.

You might be interested in

A thin plastic rod of length 2.5 m is rubbed all over with wool, and acquires a charge of 75 nC, distributed uniformly over its

Colt1911 [192]

Answer:

find out the le

Explanation:

3 0

2 years ago

PLZZZZ HELP QUICK . 30 points. Emma weighs 560 N, she has decided to stay in shape so she is working out every morning, This mor

Phantasy [73]

Force equals mass time acceleration. Weight is a force and it can replace force in the equation. The acceleration would be gravity, which is an acceleration.

1.)

Fw (weight) = m (mass) · g (gravity, 9.8 m/s²)

Fw = m * 9.81 m/s²

560N = m · 9.81 m/s²

m ≈ 57.08 kg

2.)

d = 350 meters

t = 65 seconds

velocity = d/t

velocity = 350 meters / 65 seconds

velocity ≈ 5.38 meters/sec

3.)

Force = 35N

Distance = 2 meters

Work = Force · Distance

Work = 35N · 2 meters

Work = 70 J

3 0

2 years ago

I don't get the flat line on a graph. on a distance/speed or motion graph, they say it is not moving. that is what I don't get i

timofeeve [1]

A flat line means the the speed is the same . Its moving at the same pace.

4 0

3 years ago

What is the density of a block of wood with a mass of 120 g and a volume of 200 cm

creativ13 [48]

D=m/V therefore the answer is 120/200 or 0.6

7 0

2 years ago

Two parallel conducting plates are separated by 9.2 cm, and one of them is taken to be at a potential of zero volts.What is the

True [87]

Answer:

$E=54V/cm$

$\Delta V=496.8V$ between the plates.

Explanation:

The equation for change of voltage between two points separated a distance d inside parallel conducting plates (<em>which have between them constant electric field</em>) is:

$\Delta V=Ed$

So to calculate our electric field strength we use the fact that the potential 8.8 cm from the zero volt plate is 475 V:

$E=\frac{\Delta V}{d}=\frac{475 V}{8.8cm}=54V/cm$

And we use the fact that the plates are 9.2cm apart to calculate the voltage between them:

$\Delta V=Ed=(54V/cm)(9.2cm)=496.8V$

8 0

2 years ago