Answer:
(a) 0.177 m
(b) 16.491 s
(c) 25 cycles
Explanation:
(a)
Distance between the maximum and the minimum of the wave = 2A ............ Equation 1
Where A = amplitude of the wave.
Given: A = 0.0885 m,
Distance between the maximum and the minimum of the wave = (2×0.0885) m
Distance between the maximum and the minimum of the wave = 0.177 m.
(b)
T = 1/f ...................... Equation 2.
Where T = period, f = frequency.
Given: f = 4.31 Hz
T = 1/4.31
T = 0.23 s.
If 1 cycle pass through the stationary observer for 0.23 s.
Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.
= 16.491 s.
(c)
If 1.21 m contains 1 cycle,
Then, 30.7 m will contain (30.7×1)/1.21
= 25.37 cycles
Approximately 25 cycles.
Answer:
<u><em>Circular motion requires a net inward or "centripetal" force. Without a net centripetal force, an object cannot travel in circular motion. In fact, if the forces are balanced, then an object in motion continues in motion in a straight line at constant speed.</em></u>
Explanation:
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
The impact of the material
type with which the slope is made affects the acceleration. Acceleration will
be higher and smoother if the material of the slope surface is smoother as
opposed to a texture which is not smooth. Smoother surface allows more acceleration
because it will have less friction and resistance. Otherwise the friction will
slow the object down for example a grassy ground will have more friction than a
well maintained marble floor.
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴



To find the constant k, we examine the total charge Q which is:


∴



Thus;




Hence, from equation (1), if k = 


To verify the units:

↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q



since 
