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Valentin [98]
2 years ago
6

Please answer asap, ill mark you as the brainliest

Physics
1 answer:
mina [271]2 years ago
4 0

Answer:

We can calculate the force or pull of gravity of the planets in-universe. We can also determine the trajectory of astronomical bodies and to predict their motion. It pulls all the object towards the earth. The motion of planets around the Sun.

<h2>Please mark me as brainliest</h2>

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Continents affect the directions of currents true or false
sladkih [1.3K]
True is the correct answer
7 0
4 years ago
What would the answer be​
wariber [46]

Answer:

D

Explanation:

The density has a formula of d = m / V.

The volumes are given as the same.

The density will be

d_lead = 113/ V

d_aluminum = 27/V

It does not matter for this question what the volume actually is. Just note that the value of the volume is the same for each ball.

A: incorrect. If the volumes are the same, the density is not the same because the masses are different.

B: incorrect. If the density of the two balls was less than water, the two balls would float. Never going to happen unless the balls are thin shells.

C: incorrect. The exact opposite is the answer.

D: Correct. This is the answer.

6 0
3 years ago
Which trait shared by dolphins and bats possibly lead to the evolution of echolocation in these two animal groups?
s344n2d4d5 [400]
The need to quickly move through dark environments.  <span />
6 0
3 years ago
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A sprinter accelerates from rest to 10.0 m/s in 1.35 seconds. What is their<br> acceleration?
ipn [44]

Answer:

0.135

Explanation:

You need to divide

4 0
4 years ago
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If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

8 0
3 years ago
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