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3 years ago

BRAINIEST BRAINIEST BRAINIEST BRAINIEST!!

Physics

2 answers:

hodyreva [135]3 years ago

7 0

Answer:

20m/s

Explanation:

velocity=displacement/time

velocity=60/3

velocity=20m/s

stellarik [79]3 years ago

5 0

20 m/s. Hope this helps! Credit to the guy/girl above me

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A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner an

aleksley [76]

Answer:

1) The charge on the outer shell is +4·Q

2) The charge on the inner shell is +Q

Explanation:

1) The given parameters of the spherical shell are;

The net charge on the spherical shell = 3·Q

The point charge surrounded by the spherical shell = -Q

Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;

The net charge, 3·Q = -q + x

∴ x = 3·Q + Q = 4·Q

The charge on the outer shell, x = 4·Q

2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;

-Q + y = 0

∴ y = +Q

The charge on the inner shell, y = +Q

5 0

3 years ago

Find the speed for Barkely, the dog, if he runs 5.5 miles in 3 hours

amm1812

Answer: B

Explanation:

5.5/3

6 0

3 years ago

How do laws differ from theories?

Strike441 [17]

Answer:

A theory is an explanation for what has been shown many times. A scientific law is a relationship in nature that has been proved many times and there are no exceptions.

<h3><u>PLEASE</u><u> MARK</u><u> ME</u><u> BRAINLIEST</u></h3>

6 0

2 years ago

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In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

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2 years ago

How far did an airplane go if it traveled at 700 mph for 3.5 hours?

Scrat [10]

2,450 miles. you have to do 700•3.5= 2,450

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3 years ago

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