From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)
Distance traveled is
d = 100 km = 10⁵ m
Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s
Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s
The difference in time is
302.12 - 19.49 = 282.63 s
Answer: 283 s (nearest second)
Answer:
The charge on each plate is 0.0048 nC
Explanation:
for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:
C = (A×ε)/d
=[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)
= 7.86×10^-14 F
then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:
q = C×V
= (7.86×10^-14)×(61)
= 4.80×10^-14 C
≈ 0.0048 nC
Therefore, the charge on each plate is 0.0048 nC.
The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.
<h3>
What is the tension in the cord?</h3>
The tension in the cord is calculated as follows;
T = ma + mg
where;
- a is the acceleration of the block
- g is acceleration due to gravity
- m is mass of the block
T = m(a + g)
T = 1.5(a + 9.8)
T = 1.5a + 14.7
Thus, the tension in the cord is (1.5a + 14.7) N.
If the block is at rest, the tension is 14.7 N.
<h3>Force of the force</h3>
The force with which the cord pulls is equal to the tension in the cord
F = T = m(a + g)
F = (1.5a + 14.7) N
If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.
Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.
Learn more about tension here: brainly.com/question/187404
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It doesn't matter. If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
potential energy you had at the top, no matter what kind
of route you took getting down.
___________________________
The only way I can think of that it would make a difference
would be if the shallow slide were REALLY REALLY long,
and you didn't have anything to eat all the way down.
Then you might lose some weight while you're on the slide,
and your mass might be less at the bottom than it was at the
top. Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.
But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.
The answer is calories !!
