The earliest radio broadcasts on Earth were emitted about 100 years ago. Approximately where are these initial radio waves now?
1 answer:
Answer:
100 ly are
Explanation:
The speed of light is, by definition (we define this and derive a definition of distance from there nowadays), c=299792458m/s. We want to know, at this speed, how much distance the radio signals travel in 100 years. Since each year has 365 days (not a leap one though), each day has 24 hours, each hour has 60 minutes and each minute has 60 seconds, the number of seconds in a year will be (365)(24)(60)(60)=31536000, so the distance traveled by the waves in 100 years will be:
, which, of course, are 100 light years.
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Given that,
Mass of object = 9.90 kg
Time =5.40 s
Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.
We need to calculate the displacement
Using formula of displacement
Where, = initial position
= final position
Put the value into the formula
(a). We need to calculate the work done on the object
Using formula of work done
Put the value into the formula
(b). We need to calculate the average power due to the force during that interval
Using formula of power
Where, P = power
W = work
t = time
Put the value into the formula
(c). We need to calculate the angle between vectors
Using formula of angle
Put the value into the formula
Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.
(b). The average power due to the force during that interval is 7.71 Watt.
(c). The angle between vectors is 79.7°
Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.