Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C
Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference
As we know the capacity is value that depend on the geometry of the capacitor, in our case two concentric spheres.
So Potential Difference between the spheres is given by:
ΔV=-
Where E = k*Q/ r^2
so we have 
then
Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)
Finally using C=Q/ΔV=ab/(k(b-a))
To caclulate the electric firld we first obtain the charge
Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C
so E=KQ/r^2 for both values of r
r=12.8 cm ( in meters)
r2=14.7 cm
E(r1)=4.83* 10^3 N/C
E(r2)=3.66 *10^3 N/C
Answer:
-1.03 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².
Mathematically, acceleration is expressed as
a = (v-u)/t ........................ Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time.
Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.
Substituting into equation 2
a = (7.20-13.60)/6.2
a = -6.4/6.2
a = -1.03 m/s²
Note: a is negative because, the hockey puck is decelerating.
Hence the average acceleration = -1.03 m/s²
Answer:
Acceleration = 1.428m/s2
Tension = 102.85N
Explanation:
The detailed solution is attached
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
Answer:
Explanation:
b) Gravity reduces the initial upward velocity to zero in a time of
t = v/g = 40/10 = 4 s
a) h = v₀t + ½gt² = 40(4) + ½(-10)4² = 80 m
or
v² = u² + 2as
h = (0² - 40²) / 2(-10) = 80 m
