V=IR therefore I=V/R=10/50=0.2A therefore the current is 0.2 A
Answer:
The observer sees the space-probe 9.055m long.
Explanation:
Let
be the length of the space-probe when measured at rest, and
be its length as observed by an observer moving at velocity
, then

Now, we know that
and
, and putting these into
we get:


Thus, an observer moving at 0.95c observes the space-probe to be 9.055m long.
Answer:
the initial velocity of the ball is 104.67 m/s.
Explanation:
Given;
angle of projection, θ = 60⁰
time of flight, T = 18.5 s
let the initial velocity of the ball, = u
The time of flight is given as;

Therefore, the initial velocity of the ball is 104.67 m/s.
Answer: The mass of the sculpture is 11.8kg
Explanation:
Using the equation of fundamental frequency of a taut string.
f = (1/2L)*√(T/μ) .... (Eqn1)
Where
f= frequency in Hertz =80Hz
T = Tension in the string = Mg
M represent the mass of the substance (sculpture) =?
g= 9.8m/s^2
L= Length of the string=90cm=0.9m
μ= mass density = mass of string /Length of string
mass of string =5g=0.005kg
L=0.9m
μ=0.005/0.9 = 0.0056kg/m
Using (Eqn1)
80= 1/(2*0.9) √(T/0.0056)
144= √(T/0.0056)
Square both sides
20736= T/0.0056
T= 116.12N
Recall that T =Mg
116.12= M * 9.8
M=116.12/9.8
M= 11.8kg
Therefore the mass of the sculpture is 11.8kg