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Marrrta [24]
3 years ago
11

"Two identical positive charges exert a repulsive force of 6.7 × 10−9 N when separated by a distance 3.5 × 10−10 m. Calculate th

e charge of each. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of C."
Physics
1 answer:
Tamiku [17]3 years ago
8 0

Answer:

The charge of each charge is 3.02\times10^{-19} C

Explanation:

When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:

F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} (1)

with q1 and q2 the charges, r the distance between them and k the Coulomb's constant (k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})

Because the charges we're dealing are identical positive q1=q2, then (1) is:

F_{e}=k\frac{\mid q^2 \mid}{r^{2}}

Using the values the problem give us:

6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}

solving for q:

q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}

q= 3.02\times10^{-19} C

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A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2rad/s2 until it reaches its fina
Otrada [13]

Answer:

962 rpm.

Explanation:

given,

angular acceleration = 190 rad/s²

initial angular speed = 0 rad/s

final angular speed = 7200 rpm

                                 =7200\times\dfrac{2\pi}{60}

                                 =754\ rad/s

we need to calculate the revolution of disk after 10 s.

time taken to reach the final angular velocity

    using equation of angular motion

 \omega_f - \omega_i = \alpha t

 754 - 0 =190\times t

    t = 4 s

rotation of wheel in 4 s

\theta =\omega_i t+  \dfrac{1}{2}\alpha t

\theta = \dfrac{1}{2}\alpha t^2

\theta = \dfrac{1}{2}\times 190 \times 4^2

 θ = 1520 rad

 \theta = \dfrac{1520}{2\pi}

 \theta =242\ rev

now, revolution of the disk in next 6 s

angular velocity is constant

\omega_f = \dfrac{\theta_f-\theta_i}{t_f-t_i}

754 = \dfrac{\theta_f-1520}{10-4}

θ_f = 6044 rad

θ_f = \dfrac{6044}{2\pi}

revolution of the computer hard disk

θ_f =  962 rpm.

total revolution of the computer disk after 10 s is equal to 962 rpm.

3 0
3 years ago
two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
4 years ago
According to Boyles’ law, PV = constant. If a graph is plotted with the pressure P against the volume V, the graph would be a(n)
Hunter-Best [27]

Answer:

C. hyperbola

Explanation:

From Boyle's law:

PV = k, where k is a constant

Solving for P:

P = k / V

At first glance, this equation doesn't fit any of the options.  But when you graph it, you can see that it's actually a <em>rotated</em> hyperbola.

4 0
3 years ago
Read 2 more answers
A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
I think you said everything but the answer.
jeka94

Answer:

What does that even mean?

Explanation:

7 0
3 years ago
Read 2 more answers
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