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sweet-ann [11.9K]
3 years ago
9

A dolphin emits a sound wave that hits a target 120 m away. The wave reflects back from the target to the dolphin. If the Bulk m

odulus of seawater is 2.3 x 10° N/m2 and the density of seawater is 1022 kg/m3. How long does it take the sound wave for the round trip?
Physics
1 answer:
Veronika [31]3 years ago
3 0

Answer:

The time taken for the sound wave to make the round trip is 0.16 s.

Explanation:

Given;

distance traveled by the sound wave, d = 120 m

bulk modulus of sea water, B = 2.3 x 10⁹ N/m²

density of sea water, ρ = 1022 kg

The speed of the wave is given by;

v = \sqrt{\frac{B}{\rho} } \\\\v = \sqrt{\frac{2.3*10^9}{1022} }\\\\v = 1500.16 \ m/s

Speed is given by;

Speed = \frac{Distance}{Time}

total distance of the round trip = 2 x 120m = 240 m

Time taken for the sound wave to make the round trip is given by;

Time = \frac{Distance}{Speed} \\\\Time = \frac{240}{1500.16} \\\\Time = 0.16 \ second

Therefore, the time taken for the sound wave to make the round trip is 0.16 s.

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How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?
lesantik [10]

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

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Put all the values,

P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W

So, the required power is 70 W.

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2 years ago
A car is accelerating at 30 m/s2, if the car is 400 kg how much force
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It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .

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6 0
3 years ago
Increasing the amount of current that flows through a wire ______ the strength of an electromagnet
Anestetic [448]

Electromagnet is in form of solenoid

and the magnetic field due to solenoid is given as

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
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