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3 years ago

14

A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of

the electric field at x = 0.200m on the x-axis.

1 answer:

Damm [24]3 years ago

3 0

You have to reduce 2.00 an5.00 I order to use the×that=0.800

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Why does a purple flower appear purple when white light shines on it?

grandymaker [24]

The flower absorbs all light but purple, making it appear, purple!

5 0

3 years ago

Studying neutrinos helped to explain how our Sun works but led to changes in theories of particle physics, how is this process c

julsineya [31]

Explanation:

Yes, evidently, the process is consistent with the scientific process because in scientific process falsification and modification are two very important traits. So this new concept have modified the existing theories.

Through the modification a theory is adjusted without undermining other discovering made through the theory's prediction.

There are many other evidences that prove this fact for example Einstien's Theory of relativity also changes the existing concepts.

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2 years ago

A train reaches a speed of 35.0 m/s after accelerating at a rate of 5.00 m/s2 over a distance of 40.0 m. What was the train’s in

MatroZZZ [7]

Answer:

Initial velocity, U = 28.73m/s

Explanation:

Given the following data;

Final velocity, V = 35m/s

Acceleration, a = 5m/s²

Distance, S = 40m

To find the initial velocity (U), we would use the third equation of motion.

V² = U² + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

35² = U + 2*5*40

1225 = U² + 400

U² = 1225 - 400

U² = 825

Taking the square root of both sides, we have;

Initial velocity, U = 28.73m/s

5 0

3 years ago

PLZ ANSWER!!! QUICKLY

grandymaker [24]

Light because they need light to grow

3 0

3 years ago

Read 2 more answers

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago