Answer:
D. The objects particles move faster
Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 ×
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 ×
)²×ω²× 34.64 ) / 2
0.050 ×(6.43 ×
)²×ω²× 34.64 = 108
ω² = 108 / 7.160 ×
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
The average dissipated power in a resistor in a ac circuit is:

where R is the resistance, and

is the root mean square current, defined as

where

is the peak value of the current. Substituting the second formula into the first one, we find

and if we re-arrange this formula and use the data of the problem, we can find the value of the peak current I0:
Answer:
15.75 m
Explanation:
First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.
Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.