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Greeley [361]
2 years ago
11

A ship's fathometer (an echo sounder) transmits a sound pulse and records the return of an echo 7.4 seconds later. If the speed

of sound in water is 1500 m/second, what is the water depth in meters
Physics
1 answer:
ira [324]2 years ago
6 0

Based on the velocity of sound in water,  the depth of the water is 5550 m.

<h3>What is the depth of the water if a transmitted sound pulse returns  an echo 7.4 seconds later?</h3>

The velocity of sound in water =  1500 m/s

Time taken to receive the echo = 7.4 seconds

  • Distance travelled by sound = speed * time

Distance = 1500 * 7.4

Distance = 11100 m

Since this is a to and fro distance covered

  • Depth of the water = distance travelled/ 2

Depth of water = 11100/2

Depth of water = 5550 m

Therefore, the depth of the water is 5550 m.

Learn more about echo and sound at: brainly.com/question/14090821

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Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
2 years ago
A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force
Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0
3 years ago
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A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)
just olya [345]

Answer:

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Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

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weeeeeb [17]

Answer:

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Explanation:

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Elodia [21]

Answer:

What is the pressure transmitted in the liquid on a hydraulic pump where an elephant with a weight of 40 000 N is placed on top of the large piston with an area of 40 m2. The small piston area is 4 m2.

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