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snow_lady [41]
3 years ago
13

A particle of charge of +3.40 ✕ 10−6 C is 17.5 cm distant from a second particle of charge of −2.00 ✕ 10−6 C. Calculate the magn

itude of the electrostatic force between the particles.
Physics
1 answer:
Karolina [17]3 years ago
5 0

Answer:

0.47 N

Explanation:

q1 = 3.4 x 10^-6 C

q2 = 2 x 10^-6 C

d = 17.5 cm = 0.175 m

The electrostatic force is given by

F=\frac{Kq_{1}q_{2}}{d^{2}}

F=\frac{9\times10^{9}\times3.4\times 10^{-6}\times 2\times 10^{-6}}{0.175\times 0.175}

F = 0.47 N

Thus, the force is 0.47 N.

You might be interested in
What is the velocity of a wave with a wavelength of 9 meters and a period of 0.006
Ipatiy [6.2K]

Answer: 1,500m/s

Explanation:

Relationship existing between velocity of a wave (v), wavelength(¶) and frequency(f) is

v = f¶... (1)

Since Frequency (f) is the reciprocal of the period (T);

Frequency = 1/Period i.e F = 1/T... (2)

Substituting equation 2 into 1 we have;

v = 1/T × ¶

v = ¶/T

Given wavelength ¶ = 9m

Period T = 0.006s

v = 9/0.006

v = 1,500m/s

The velocity of the wave will be 1,500m/s

8 0
3 years ago
Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency of 440.0 hz. The oth
Katyanochek1 [597]

Answer:

Percentage change in tension is 3.8%

Explanation:

We have given initially frequency f_1 = 440 Hz

Let tension in the string at this frequency is T_1

Now second frequency is f_2=448.4Hz

Frequency in string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}

From the relation we can say that

\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}

\frac{440}{448.4}=\sqrt{\frac{T_1}{T_2}}

{\frac{T_2}{T_1}}=1.038

Percentage change in tension is equal to

=\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=(1.038-1)\times 100=3.8 %

So percentage change in tension is 3.8%

4 0
3 years ago
WILL GIVE BRAINLYST A flower pot is thrown out of a window with a horizontal velocity of 8 m/s. If the window is 1.5 m off the g
stich3 [128]
-- Use the formula  D=(1/2)·(g)·(t²)  to calculate how long
it takes the flower pot to fall to the ground.

1.5 m = (1/2) · (9.8 m/s²) · (t²)

t² = (1.5 m) / (4.9 m/s²)

t = 0.554 second

-- Use the formula Distance = (speed)·(time) to calculate
how far the pot traveled horizontally in that amount of time.

Distance = (8 m/s) · (0.554 sec)

Distance horizontally = 4.43 meters
3 0
3 years ago
Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He
umka2103 [35]
<h3>Answer:  130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

  • The initial speed (let's call that Vi) is 8.0 m/s
  • The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
  • This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

  • At 0 seconds, he's going 8 m/s
  • At the 1 second mark, he's slowing down to 8-2 = 6 m/s
  • At the 2 second mark, he's now at 6-2 = 4 m/s
  • At the 3 second mark, he's at 4-2 = 2 m/s
  • Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

  • F = force applied
  • m = mass
  • a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0
3 years ago
4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
KatRina [158]

Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

8 0
3 years ago
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