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Vedmedyk [2.9K]
3 years ago
7

Defination coulomb's law.

Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

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Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.70 m/s. Her husband Bruce suddenly realiz
Katen [24]

Answer:

The velocity is v  =  6.66 \  m/s

Henrietta is at distance s=  18.17 \  m from the under the window

Explanation:

From the question we are told that

The speed of Henrietta is v=  2.70 \ m/s

The height of the window from the ground is h  =  36.5 \  m

Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is

t  =  \sqrt{\frac{2 *  h }{g} }

=> t  =  \sqrt{\frac{2 *  36.5 }{9,8} }

=> t  =  2.73 \  s

Generally the time taken for the lunch to reach Henrietta is mathematically represented as

T =  t +  t_1

Here t_1 is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

Now

T = 2.73  +  4

=> T = 6.73 \  s

Generally the distance covered by Henrietta before catching her lunch is

s=  v  *  T

=> s=  2.70  * 6.73

=> s=  18.17 \  m

Generally the speed with which Bruce threw her lunch is mathematically represented as

v  =  \frac{18.17}{2.73}

v  =  6.66 \  m/s

4 0
3 years ago
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it
fomenos

Answer:68.15m/s

Explanation:

<u><em>Given: </em></u>

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

<u><em>Formula:</em></u>

v₁²=v₁²+2a (x)

<u>Set up:</u>

=\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)

<h2><u><em>Solution:</em></u></h2><h2><u><em>68.15m/s</em></u></h2>

<u />

6 0
3 years ago
What is the accletation of a 1500kg with a net force of 7500 N
matrenka [14]
Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.
3 0
3 years ago
A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration
Tpy6a [65]

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

v^2 = u^2 + 2as

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 m/s^2

s = 91 m

Therefore:

v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s

The velocity of the boat when it reaches the buoy is 7.5 m/s.

6 0
2 years ago
Describe the relationship between a moving objects mass and its kinetic energy
swat32
Kinetic energy is the energy of mass in motion, the kinetic energy that an object has is because of it's motion.Heavier objects that are moving have more kinetic energy than lighter ones. 
6 0
3 years ago
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