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djyliett [7]
3 years ago
15

A 4.0 kg shot put is thrown with 30 N of force. What is its acceleration?

Physics
1 answer:
weeeeeb [17]3 years ago
7 0

7.5 m/s

a = F ÷ m

a = (30 N) ÷ (4.0 kg)

a = 7.5 m/s2

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Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri
9966 [12]

Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

F = 5.625 * 10^{-4} N

Now in order to find the acceleration of each mass

we can use

F = ma

5.625 * 10^{-4} = 1 * a

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8 0
3 years ago
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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = \sqrt{2I/ce_{0} }

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9}  = <em>1.55 x 10^-8 v/m</em>

6 0
3 years ago
A fighter jet accelerates from 10 m/s to 75 m/s in 7.0 s. The acceleration of the jet is
liraira [26]
Acceleration = final velocity - inital / time
a = 75-10 / 7
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4 0
2 years ago
Help me please!! 15points
Romashka-Z-Leto [24]

Explanation:

Acceleration is the change in speed over change in time.

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a = 2.1 m/s²

c. The stone's speed is:

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4 0
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The answer that correctly describes the grey lines in this Hering illusion is "The grey lines are bent." Option A. This is further explained below

<h3>What is the Hering illusion?</h3>

The Hering Illusion is one of several illusions in which a major component of a basic line picture is obscured.

In conclusion, The statement for the grey lines in this Hering illusion is "The grey lines are twisted.".

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