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Orlov [11]
3 years ago
13

What is the trend noticed with the inner planets

Physics
1 answer:
lutik1710 [3]3 years ago
7 0
The inner planets are closer to the Sun and are smaller and rockier. ... The outer planets are further away, larger and made up mostly of gas. The inner planets (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars.Apr 23, 2014
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Explain how a bathroom scale is like a biofeedback machine.
amid [387]

Answer:

                 

Explanation:

4 0
3 years ago
The purpose of striking the ball in a volleyball game is to: 1. Place the ball in motion 2. Change the direction of the ball's m
zhenek [66]

Answer:

4. All of the above

Explanation:

The purpose of striking the ball in a volleyball game:

From the serve you could state that you need to place the ball in motion.

When returning a shot of, you normally want to change the direction of the ball's motion.

During a dropshot, you purposely want to slow down the ball's motion.

The correct answer must be all of the above.

8 0
3 years ago
The resistivity of water is increased if salt is added. true or false
sergejj [24]
The answer is true hopes I helped
3 0
3 years ago
Read 2 more answers
Calculate the potential energy of a body of a mass 2kg held 4 meters above the ground if g=10m/s?​
Drupady [299]

Answer:

16000

Explanation:

Mass(m)=2Kg (1kg= 1ooo g then 2 kg=2000 g)

Velocity(v)= 4 meter

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

We know that,

P.E= 1/2 mv^2

or, 1/2 × 2ooo × 4^2

or, 1/2×2000 ×16

or, 2000×8

Therefore= 16000

7 0
2 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
Read 2 more answers
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