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Ainat [17]
2 years ago
13

A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th

e cube's sides is -1 kN·m^2/C. How much charge is at the center?
Physics
1 answer:
12345 [234]2 years ago
5 0

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, L=10\ cm

The Elecric flux through one of the side of the cube is, \phi =-1 kNm^2/C.

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by \epsilon_0

Since Flux is a scalar quantity. It can added to get total flux through the surface.

\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C

So the the charge at the centre is calculated.

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A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain
Mama L [17]

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.  

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

4 0
2 years ago
Aiden takes 0.10s to throw a baseball, which leaves his hand with a velocity of 49m/s. The balls acceleration is
Brums [2.3K]
Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
a =  \frac{v}{t}
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
a =  \frac{49}{0.1}  \\ a =  \frac{490}{1}  \\ a = 490
The acceleration is 490 m/s^2
7 0
3 years ago
A drag racer starts her car from rest and accelerates at 5.5 m/s2 for the entire distance of 523 m. How long did it take the car
ANEK [815]

Answer:

t = 13.7 s or t = 14 s with proper significant figures

Explanation:

The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.

Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.

d = vt + \frac{1}{2}at²

523 = (0)t + (\frac{1}{2})(5.5)t²

523 = 2.8t²

186.8 = t²

13.7 s = t

(t = 14 s with proper significant figures)

3 0
2 years ago
In a double‑slit interference experiment, the wavelength is lambda=487 nm , the slit separation is d=0.200 mm , and the screen i
pantera1 [17]

Answer:

Δx = 4.68 x 10⁻³ m = 4.68 mm

Explanation:

The distance between the consecutive maxima, in Young's Double Slit Experiment is given bu the following formula:

Δx = λD/d

So, the distance between the eighth order maximum and the fourth order maximum on the screen will be given as:

Δx = 4λD/d

where,

Δx = distance between eighth order maximum and fourth order maximum=?

λ = wavelength = 487 nm = 4.87 x 10⁻⁷ m

d = slit separation = 0.2 mm = 2 x 10⁻⁴ m

D = Distance between slits and screen = 48 cm = 0.48 m

Therefore,

Δx = (4)(4.87 x 10⁻⁷ m)(0.48 m)/(2 x 10⁻⁴ m)

<u>Δx = 4.68 x 10⁻³ m = 4.68 mm</u>

5 0
3 years ago
A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hol
il63 [147K]

Answer:

v = 12.1 m/s

Explanation:

  • When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.
  • These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.
  • In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.
  • Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.
  • This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.
  • At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):

       F_{c} = N + m*g (1)

  • Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:

       F_{c} = m*\frac{v^{2}}{r} (2)

  • Since the normal force takes any value as needed to make (1) equal to (2),  if the speed diminishes, it will be needed less force to keep the equality valid.
  • In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.
  • In this condition, from (1) and (2), we can find the minimum possible value of  the speed that still keeps the motorcycle touching the surface, as follows:
  • v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)
6 0
3 years ago
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