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3 years ago

Which of the following is an example of mechanical waves?

Physics

1 answer:

Nastasia [14]3 years ago

8 0

Sound Waves will be an example of mechanical waves.. hope this helps!

You might be interested in

What is inertia and how is it related to the newton's first law of motion?

Alina [70]

Answer:

Inertia is the resistance of any physical object to any change in its velocity. This includes changes to the object's speed, or direction of motion. An aspect of this property is the tendency of objects to keep moving in a straight line at a constant speed, when no forces act upon them.

Explanation:

Some sort of a local field, maybe not our A field, is really the cause of inertia. When you push on an object a gravitational disturbance goes propagating off into either the past or the future. Out there in the past or future the disturbance makes the distant matter in the universe wiggle.

7 0

3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

$f_r = 150.47 N$

8 0

3 years ago

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

4 years ago

a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.

Allushta [10]

Answer:

=3 metre per second ^2

Explanation:

Formula for acceleration is

V-U÷T

In the given information

V=16

U=4

T=4

Acceleration =16-4/4

=3 metre per second ^2

4 0

3 years ago

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

love history [14]

Answer:

$ans \: = \boxed{{4.8 \times 10}^{ - 4} Hz}$

Explanation:

$given \to \\ f_{r} = 350 \: \\ v_{r} = {3 \times 10}^{8} \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{r} = \frac{v_{r}}{f_{r}} = \frac{3 \times 10^{8} }{350} = \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\ f_{w} = water \: frequency = \: \boxed{ ?}\: \\ v_{w} = 14 50 \\ but \to \\ v = f \gamma \to \: \gamma = \frac{v}{f} : hence \to \\ \gamma _{w} = \frac{v_{w}}{f_{w}} = \frac{1}{100} \times \gamma _{r} = \frac{1}{100} \times 857,142.85714 \\\gamma _{w} = \boxed{8,571.4285714 \: m} : hence \to \: \\ f_{w} = \frac{v_{w}}{ \gamma _{w}} = \frac{1450}{8,571.4285714} = \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) = \frac{f_{w}}{f_{r}} = \frac{0.1691666667}{350} = 0.0004833333 \\ hence \to \\ the \: frequency \: of \: the \: radio \: wave \: is \to \: \boxed{{4.8 \times 10}^{ - 4} }\: \\ that \: of \: the \: wave \: created \: in \: the \: water.$

♨Rage♨

8 0

3 years ago