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sweet [91]
3 years ago
5

Two soccer players kick the same 1-kg ball at the same time in opposite directions. One kicks with a force of 25 N; the other ki

cks with a force of 10 N. What is the resulting acceleration of the ball, in m/s2 Answer here I​
Physics
2 answers:
Studentka2010 [4]3 years ago
7 0

Answer:

15 (These are just here to increase the word limit)

andre [41]3 years ago
5 0

Answer:

The forces offset to produce a net force of 8 N

since the ball is 1 kg

F = ma

a = f/m = 8 N/1 kg = 8 (kg m)/(s^2 kg) = 8 m/s^2

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Sliding friction between two objects that are moving past each other can be increased by: a. putting sand between the objects b.
alukav5142 [94]

Answer:

Putting sand between the objects

Explanation:

Friction is the action of rubbing against an object, which can POSSIBLY create static electricity.

8 0
3 years ago
Read 2 more answers
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
(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down
slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

h=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0
3 years ago
Could we see a galaxy that is 20 billion light-years away?.
vichka [17]

Answer:

No

Explanation:

20 billion light-years away are beyond our sight and perspective on Earth and wouldn't be observable in our universe.

7 0
2 years ago
Thomas needs to move an 80 kg rock, but cannot lift it. He decides to use a
ArbitrLikvidat [17]

Answer:

4

Explanation:

The weight of the rock is W = mg = (80 kg) (10 m/s²) = 800 N.

The mechanical advantage is therefore 800 N / 200 N = 4.

5 0
3 years ago
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