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meriva
4 years ago
14

As shown in the diagram, threee equal charges are spaced evenly in a row. The magnitude of each charge is +2e, and the distance

between two adjacent charges is 1.50 nm. Then the central charge is displaced 0.350 nm to the right while the other two charges are held in place. After the displacement, what is the magnitude and direction of the net force that the outer two charges exert on the central charge?
4.27*10^-10 to the right

4.27*10^-10 to the left

3.20*10^-10 to the left

3.20*10^-10 to the right

Physics
2 answers:
denpristay [2]4 years ago
7 0

Answer:

4.27*10^(-10) to the left

Explanation:

Force exerted by right charge

F_{right} = \frac{k*(2e)*(2e)}{((1.5 - 0.35)*10^(-9))^2} \\\\F_{right} = \frac{8.99*10^9 * 4 * (1.6*10^(-19))^2}{((1.5 - 0.35)*10^(-9))^2} \\\\F_{right} = 6.96*10^(-10) N

Force exerted by left charge

F_{left} = \frac{k*(2e)*(2e)}{((1.5 + 0.35)*10^(-9))^2} \\\\F_{left} = \frac{8.99*10^9 * 4 * (1.6*10^(-19))^2}{((1.5 + 0.35)*10^(-9))^2} \\\\F_{left} = 2.689*10^(-10) N

Resultant Force

F_{res} =  F_{right} - F_{left}\\F_{res} = 6.96*10^(-10) - 2.689*10^(-10)\\\\F_{res} = 4.271 * 10 ^(-10) N

Hence, right charge exerts more force than left so central experience the above force in left direction.

Nataly [62]4 years ago
4 0

Answer: 4.27 x 10^-10 N to the left

Explanation: I just took this quiz

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german
<h2>Hello!</h2>

The answer is:

The first option,  the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

Distance=NorthCoveredDistance+EastCoveredDistance

Distance=4*180m+3*180m=720m+540m=1260m

Actual distance between the start and the end point (displacement):

ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0
3 years ago
At standard temperature and pressure, carbon dioxide has a density of 1.98 kg/m3. What volume does 1.70 kg of carbon dioxide occ
nikitadnepr [17]

Answer:

<h2>volume= 0.85m^3</h2>

Explanation:

<em>The density of a substance is defined as the mass per unit volume of the substance, the unit is in kg/m^3 and it is represented by the greek letter rho</em>

Step one:

given data

we are told that the density  of Co2=  1.98 kg/m3

and the mass of Co2 is= 1.70 kg

we know the relation between mass, volume and density is

density=mass/volume

make volume subject of formula we have

volume=mass/density

substitute we have

volume=1.7/1.98\\\\volume= 0.85m^3

8 0
3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car
kogti [31]

Answer:

t_1 = \frac{v_i}{a_i}

t_2 = \frac{v_i}{a_i}

Δd = v_it_1 = v_i^2/a_i

Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

8 0
3 years ago
How long does it take light to travel 850 km in a vacuum? Answer in ms.(Express your answer to two significant figures.)
poizon [28]

Answer:

0.002833 sec

Explanation:

Speed of light in vacuum is 3\times 10^{8}m/sec

Given distance = 850 km = 850×1000=850000 m

We have to calculate the time that light take to travel the distance 850 km

Time T=\frac{distance }{speed}=\frac{850000}{3\times 10^8}=2.833\times 10^{-3}sec

So the time taken by light to travel 850 km is 0.002833 sec

5 0
3 years ago
The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
choli [55]

Answer:

25 times the average speed

7 1
3 years ago
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