On a warm summer day (31 ∘c), it takes 4.60 s for an echo to return from a cliff across a lake. on a winter day, it takes 5.00 s
1 answer:
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.
First of all, we need to calculate the speed of sound at temperature of
:
The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
First of all, we need to calculate the speed of sound at temperature of
The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
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Answer:
(a) Magnetic moment will be
(b) Torque will be
Explanation:
We have given dimension of the rectangular 5.4 cm × 8.5 cm
So area of the rectangular coil
Current is given as
Number of turns N = 25
(A) We know that magnetic moment is given by
(b) Magnetic field is given as B = 0.350 T
We know that torque is given by
a) Average power: 1425 W
b) Instantaneous power at 3.0 sec: 2850 W
Explanation:
a)
The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation
where
s = 18 m is the displacement along the ramp
u = 0 is the initial velocity
t = 3.0 s is the time taken
a is the acceleration of the object along the ramp
Solving for a,
Now we can apply Newton's second law to find the net force on the object:
This net force is the resultant of the applied force forward () and the component of the weight acting backward (
), so we can find what is the applied force:
where
m = 24 kg is the mass of the object
is the acceleration of gravity
Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:
where s = 18 m is the displacement.
Therefore the average power needed is:
b)
The instantaneous power at any point of the motion is given by
where
is the force applied
v is the velocity of the object
We already calculated the applied force:
While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:
And therefore, the instantaeous power at 3.0 sec is:
Learn more about power:
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