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scZoUnD [109]
4 years ago
10

What is the magnitude of the force a 25 charge exerts on a 3 mc charge 35cm away?

Physics
1 answer:
lara31 [8.8K]4 years ago
4 0
The electrostatic force between two charges is given by
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the charges

In our problem, 
q_1 = 25 \mu C=25 \cdot 10^{-6} C
q_2 = 3 mC = 3 \cdot 10^{-3} C
r=35 cm=0.35 m
Therefore the electrostatic force is
F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(25 \cdot 10^{-6}C)(3 \cdot 10^{-3}C)}{(0.35 m)^2}= 5510 N
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Grrr i don't have much time, help please T^T
mart [117]

Answer:

2. ( b ) zero

3. ( c ) 10 s

4. Uniform then decreasing

Explanation:

2.

Since the motion is uniform, initial and final velocity will be 0, hence acceleration will be zero.

3.

Initial velocity ( u ) = 5 m/s

Final velocity ( v ) = 35 m/s

Acceleration ( a ) 3 m/s^2

To find : Time ( t )

Formula : -

t = v - u / a

 = 35 - 5 / 3

 = 30 / 3

t = 10 s

4 0
2 years ago
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A man standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down the man continues
vovikov84 [41]

When the bus starts moving forward, the man remains at rest,
causing him to lean back.

When the bus slows down, the man continues to move forward,
and appears to lean forward.

Both events are examples of the effect of inertia.

6 0
4 years ago
The sides of a square increase in length at a rate of 3 ​m/sec. a. At what rate is the area of the square changing when the side
maks197457 [2]

The area of a square is given by:

A = s²

A is the square's area

s is the length of one of the square's sides

Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:

d[A]/dt = d[s²]/dt

The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt

dA/dt = 2s(ds/dt)

A) Given values:

s = 14m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

dA/dt = 2(14)(3)

dA/dt = 84m²/s

B) Given values:

s = 25m

ds/dt = 3m/s

Plug in these values and solve for dA/dt:

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dA/dt = 150m²/s

6 0
3 years ago
In a test of an energy-absorbing bumper, a 2800-lb car is driven into a barrier at 5 mi/h. The duration of the impact is 0.4 sec
Fiesta28 [93]

Answer:

F = 2074.13 lb

Explanation:

Given that,

Mass of car, m = 2800 lb = 1270.059 kg

Initial speed, u = 5 mi/h = 2.2352 m/s

Final speed, v = - 1.5 mi/h = -0.67056 m/s  (in opposite direction)

Time, t = 0.4 s

We need to find the magnitude of the average horizontal force (lb) exerted on the car during the impact. It can be calculated as :

F=m\times \dfrac{v-u}{t}\\\\F=1270.059\times \dfrac{-0.67056 -2.2352 }{0.4}\\\\F=9226.21\ N

or

F = -2074.13 lb

So, the required force is 2074.13 lb.

5 0
3 years ago
Select the volume units that are greater than one liter
dsp73

-- kiloliter ... 1000 liters

-- megaliter ... 1 million liters

Those two are the only ones on the list that are bigger than 1 liter.

The others:

-- milli- ... 0.001

-- deci- ... 0.1

-- centi- ... 0.01

-- nano- ... 0.000 000 001

6 0
3 years ago
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