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Basile [38]
3 years ago
9

Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below. How much time el

apses before the dolphin hears the echoes of the clicks? (The speed of sound in seawater is approximately 1530 m/s.)
Physics
1 answer:
umka2103 [35]3 years ago
4 0

Answer:

Time, t = 0.104 seconds

Explanation:

Frequency of the click of the Dolphin, f = 55.3 kHz

A dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below, d = 80 m

The speed of sound in seawater is, v = 1530 m/s

Once the sound is send and reflects, the total distance covered by it is 2d such that,

t=\dfrac{2d}{v}\\\\t=\dfrac{2\times 80}{1530}\\\\t=0.104\ s

So, the time elapses before the dolphin hears the echoes of the clicks is 0.104 seconds.

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when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe
Alinara [238K]

Answer:

Rate of change of area will be 9.796cm^2/min

Explanation:

We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

Radius of the circular plate r = 52 cm

Area is given by A=\pi r^2

So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

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3 years ago
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
2 years ago
A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in:
Stells [14]

A child slides down a hill on a toboggan with acceleration of 1.8 m/s2. if she starts from rest, how far has she traveled in: 2 seconds

Answer:

2.4 m

Explanation:

From the question above,

Applying equation of motion,

s = ut+at²/2....................... Equation 1

Where t = time, u = initial velocity, a = acceleration,  s = distance.

make s the subject of the equation,

Given: a = 1.8 m/s²,  t = 2 seconds, u = 0 m/s (from rest)

Substitute these value into equation 1

s = 0(2)+1.8(2²)/2

s = 1.2(4)/2

s = 2.4 m.

Hence she has traveled 2.4 m

4 0
3 years ago
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