PHYSICS COLLEGE

A point charge that is exactly q =3 mu or micro CC is at the origin. In this problem, assume that the Coulomb constant k = 8.99 times 109 N m2/C2 exactly.(a) Find the potential V on the x axis at x = 3.00 m and at x = 3.01 m. In this part, enter your answers to exactly 6 signfificant figures.

Answers

Answer 1

Answer:

Answer:

a) 8,990.00 V b) 8,960.13 V

Explanation:

a) The potential due to a point charge, can be found from the expression of Coulomb's Law, as follows:

$V = (k*q)/(r)$

where k = 8.99*10⁹ N*m²/C², q = 3.00*10⁻⁶ C, and r = 3.00 m.

Replacing by this values, we can find the potential V as follows:

$V = (8.99e9 N*m2/C2*3.00e-6C)/(3.00m) = 8,990.00 V$

b) Repeating the process for r = 3.01m:

$V = (8.99e9 N*m2/C2*3.00e-6C)/(3.01m) = 8,960.13 V$

Answer 2

Answer:

Final answer:

The potential V at x=3.00 m and x=3.01 m from a point charge at the origin is 8.99 * 10^3 V and 8.97 * 10^3 V, respectively. This calculation is based on Coulomb's Law.

Explanation:

The potential V at a distance r from a point charge q is given by Coulomb's Law:

V = k*(q/r)

Here, k = 8.99 * 10^9 N*m^2/C^2 is the Coulomb constant, q = 3 µC is the charge, and r is the distance from the origin along the x-axis. For x = 3.00 m and x = 3.01 m, we can substitute these values into the equation to find V:

V1 = ((8.99 * 10^9 N*m^2/C^2)* (3 * 10^-6 C)) / 3.00 m = 8.99 * 10^3 V (to 6 significant figures)
V2 = ((8.99 * 10^9 N*m^2/C^2)* (3 * 10^-6 C)) / 3.01 m = 8.97 * 10^3 V (to 6 significant figures)

Learn more about Electric Potential here:

brainly.com/question/32897214

#SPJ3

Related Questions

Sophia wanted to estimate the product 73 x 12 so she made an estimate of 70x 10. Would her estimate be greater or near the actual estimate?
A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.
In college softball, the distance from the pitcher's mound to the batter is 43 feet. If the ball leaves the bat at 110 mph , how much time elapses between the hit and the ball reaching the pitcher?
What is the weight of a 45 kg box?
A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

Answers

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net?

Answers

Answer:

ball clears the net

Explanation:

$v_(o)$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_(ox)$ = initial velocity = $v_(o) Cos\theta = 20 Cos5 = 19.92 ms^(-1)$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t \n7 = v_(ox) t\nt = (7)/(v_(ox))$ Eq-1

Consider the motion of the ball along the vertical direction

$v_(oy)$ = initial velocity = $v_(o) Sin\theta = 20 Sin5 = 1.74 ms^(-1)$

$t$ = time of travel

$Y_(o)$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_(y)$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_(o) + v_(oy) t + (0.5) a_(y) t^(2)\nY = 2 + (20 Sin5) ((7)/(20 Cos5)) + (0.5) (- 9.8) ((7)/(20 Cos5))^(2)\nY = 2.00758 m\n$

Since Y > 1 m

hence the ball clears the net

Average wavelength of radio waves

Answers

The average wavelength of radio waves ranges from roughly two millimeters to more than 150 kilometers. The wavelengths of radio waves are the longest in the electromagnetic spectrum

What is Wavelength?

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The wavelength is inversely proportional to the frequency of the wave as from the following relation.

C = νλ

They also have the lowest frequencies, ranging from around 4,000 cycles per second, or 3 kilohertz, to roughly 280 billion hertz, or 280 gigahertz.

The wavelengths of radio waves are the longest in the electromagnetic spectrum, ranging from roughly two millimeters to more than 150 kilometers.

To learn more about wavelength from here, refer to the link given below;

brainly.com/question/7143261

#SPJ6

Answer:

Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km

Early in the morning, when the temperature is 5.5 °C, gasoline is pumped into a car’s 53-L steel gas tank until it is filled to the top. Later in the day the temperature rises to 27 °C. Since the volume of gasoline increases more for a given temperature increase than the volume of the steel tank, gasoline will spill out of the tank. How much gasoline spills out in this case?