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trasher [3.6K]
4 years ago
5

The magnitude of the electrical force acting between a +2.4 x 10- C charge and a +1.8 x 10- C charge that are separated

Physics
1 answer:
Brrunno [24]4 years ago
3 0

Answer: 6.07 N

Explanation:

According to Coulomb's Law:  

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=2.4(10)^{-8} C and q_{2}=1.8(10)^{-6} C are the electric charges

d=0.008 m is the separation distance between the charges

Solving:

F_{E}= 8.99(10)^{9} Nm^{2}/C^{2}\frac{(2.4(10)^{-8} C)(1.8(10)^{-6} C)}{(0.008 m)^{2}}  

F_{E}=6.07 N  

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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
A shot putter shoots a 7.3kg shot from rest to 14m/s in 1.5s. what was the average power?
sweet [91]

Answer: 477W

Explanation:

Given the following :

Mass (m) = 7.3kg

Initial Velocity (u) = 0

Final velocity (v) = 14m/s

time (t) = 1.5s

Power = workdone (W) / time (t)

The workdone can be calculated as the change in kinetic energy (KE) :

Recall ;

KE = 0.5mv^2

Therefore, change in KE is given by:

0.5mv^2 - 0.5mu^2

Change in KE = 0.5(7.3)(14^2) - 0.5(7.3)(0^2)

Change in KE = 715.4J

Therefore ;

Average power = Workdone / time

Workdone = change in KE = 715.4N

Average power = 715.4 / 1.5

Average power = 476.93333 W

= 477W

5 0
4 years ago
A 6 cm long spring extends to 9 cm when a 1 kg load is suspended from it. What would be its length if a 2 kg load were suspended
Ganezh [65]
The answer is D that is the right question
8 0
3 years ago
A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg
djverab [1.8K]

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

7 0
3 years ago
Read 2 more answers
Why can someone please help me out
denis23 [38]
D I think hope this helps
4 0
3 years ago
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