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zalisa [80]
3 years ago
9

What is the uses of echoes​

Physics
1 answer:
Anestetic [448]3 years ago
6 0

Answer:

Echoes are the reflection of sound from relatively flat object that is far enough away that you can discern the time difference. Echoes are used to measure distance, velocity, and the shape of objects. Echoes off gratings result in an unusual pinging sound

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The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re
weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

5 0
2 years ago
If there were no external forces acting on the two pucks, their complex motion could be described as the combination of the unif
mote1985 [20]

The absence of external forces will make the pucks move in the form of a uniform circular motion.

<h3>What is a circular motion?</h3>

It should be noted that a circular motion simply means the movement of an object along the circumference of the circle.

In this case, the absence of external forces will make the pucks move in the form of a uniform circular motion.

If the friction is absent, the pucks will continue to move on the same path due to the first law of Newton and the law of conversation of energy. In this case,the results will match the predictions until there's loss in energy.

Learn more about circular motion on:

brainly.com/question/106339

3 0
2 years ago
When Bella pushes a box with a mass of 5.25 kilograms with a force of 15.75 newtons, it accelerates at a rate of 2.5 meters/seco
Mariana [72]

Sure !

Start with Newton's second law of motion:

                     Net Force = (mass) x (acceleration) .

This formula is so useful, and so easy, that you really
should memorize it.

Now, watch:

The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of  2.5 m/s² .
What's the net force on the box ?

                    Net Force = (mass) x (acceleration)

                                     = (5.25 kilograms) x (2.5 m/s²)

                     Net force =       13.125 newtons .

But hold up, hee haw, whoa !  Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?

==>  Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.

How much ?

            (Bella's 15.75 newtons) minus (13.125 that the box feels)

           =      2.625 newtons backwards, applied by friction.


5 0
3 years ago
Read 2 more answers
a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d
USPshnik [31]

Answer:

The time is t = 2.595 \  s

The speed is v = 25.43 \ m/s

Explanation:

From the question we are told that

    The height of the cliff is  h =  33 \  m

 Generally from kinematic equation we have that

       h  =  ut + \frac{1}{2} gt^2

before the jump the persons initial velocity is  u =  0 m/s

 So

        33   =  0 * t + \frac{1}{2} 9.8 * t^2

=>      t = 2.595 \  s

Generally from kinematic equation

     v= u + gt

=>  v= 0 + 9.8 * 2.595

=>  v = 25.43 \ m/s

3 0
3 years ago
g A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has
Nikitich [7]

Answer:

Explanation:

gravitational acceleration of meteoroid

=  GM / R²

M is mass of planet , R is radius of orbit of meteoroid from the Centre of the planet .

R = (.9 x 6370 + 600 )x 10³ m

= 6333 x 10³ m

M , mass of the planet = 5.97 x 10²⁴ kg .

gravitational acceleration of meteoroid

=  GM / R²

=  (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ kg  / (6333 x 10³ m)²

9.92m/s²

3 0
3 years ago
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