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rusak2 [61]
3 years ago
8

Two spherical objects are separated by a distance that is 9.00 X 10m The objects are initially electrically neutral and are very

small compared to the stance between them. Each objectures the same negative charge due to the addition of electrons. As a result, each experiences an lectr ic force that has a magnitude of 1.137 10 . How many electrons did t o produce the charge on one of the objects?
Physics
1 answer:
romanna [79]3 years ago
5 0

Answer: 3.2 * 10^-3 C

Explanation: In order to solve this problem we have to use the Coulomb force given by:

F=k*q^2/ d^2 where he consider the same charge for each point

so, we have

q^2= F*d^2/k= 1.137* (10 C*9*10m)^2/9*10^9 N*m^2/C^2)=3.2 * 10^-3 C.

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A 0.15 kg baseball is pushed with 100 N force. what will its acceleration be?
sammy [17]

Answer:

The acceleration of the ball is 666.67 m/s²

Explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Applied force to it, F = 100 N

We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :

F = ma

a=\dfrac{F}{m}

a=\dfrac{100\ N}{0.15\ kg}

a=666.67\ m/s^2

So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.

8 0
3 years ago
The sound intensity level from one solo flute is 70 dB. If 10 flutists standing close together play in unison, what will the sou
Masja [62]

Answer:

80 dB

Explanation:

I_0 = Threshold intensity = 10^{-12}\ W/m^2

I = Intensity of sound

\beta = Intensity level of sound = 70 dB

Intensity level of sound is given by

\beta=10log\dfrac{I}{I_0}\\\Rightarrow 70=10log\dfrac{I}{I_0}\\\Rightarrow \dfrac{70}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{70}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{I}{10^{-12}}=10^{\dfrac{70}{10}}\\\Rightarrow I=10^{7}\times 10^{-12}\\\Rightarrow I=10^{-5}\ W/m^2

If there are 10 flutes I=10\times 10^{-5}\ W/m^2

\beta=10log\dfrac{10\times 10^{-5}}{10^{-12}}\\\Rightarrow \beta=10log10^8\\\Rightarrow \beta=10\times 8\\\Rightarrow \beta=80\ dB

The sound intensity level is 80 dB

5 0
3 years ago
Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 156.9 s ( 2 min and 36.9 s ) , wherea
Leya [2.2K]

Answer:

Jill average velocity is  0

Jack  average velocity is 0.159337

Jill average speed = 1.593372

Jack average speed = 1.434034

Explanation:

given data

long swimming pool = 25.0 m

9 lengths of the pool = 156.9 s ( 2 min and 36.9 s )

10 lengths = same time interval

to find out

average velocity and average speed

solution

we know that average velocity that is express as

average velocity = \frac{displacement}{time}    .....................1

Jill come back where she start

so here velocity will be = 0

and

Jack ends up on the other end of pool

so average velocity =  \frac{25}{156.9}

average velocity = 0.159337

now we get here average speed that is express as

average speed = \frac{distance}{time}      .............2

jack speed = 9 × \frac{25}{156.9}

jack speed = 1.434034

and

Jill speed = 10 × \frac{25}{156.9}

Jill speed = 1.593372

5 0
3 years ago
Increasing temperature increases solubility so that 480 grams (g) of sugar makes a saturated solution in 100 milliliters (mL) of
timofeeve [1]
150 ml of saturated solution
5 0
3 years ago
Read 2 more answers
What is the velocity of an object that has a momentum of 4000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.
insens350 [35]
The answer is C. You divide 4000 kg/s by 115 kg.
5 0
3 years ago
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