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3 years ago

Please answer correctlywill give the brainliestUrgent

Physics

1 answer:

EastWind [94]3 years ago

3 0

Answer:

f (frequency) = V / y where V is the speed of sound and y the wavelength

f = 1500 m/s / 1.5 m = 1000 / sec

T (period) = 1 / f = .001 sec

Suppose you replace the horn by a drum then the period would be the time between the beats of the drum - now if the source is moving towards the observer then the distance between crests of the wave produced by the drum will be shortened by V * T because of the motion of the drum "towards" the observer, and since the wavelength is shorter the frequency heard by the observer will be higher, and the higher the speed of of the car the shorter the wavelength as seen by the observer and the higher the frequency.

Also, if the car is moving away from the observer then the distance between the crests of the wave emitted will be further apart, and the observer will hear a lower frequency.

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A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

For the hoop:

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly for the solid disk with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

4 0

3 years ago

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My question is attached!

Lesechka [4]

Answer:

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I need help with this one Coulombs law 100 points and brainliest

Ganezh [65]

Answer:

my dad is a physics professor so when he comes home i will tell him to answer the question for you

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3 years ago

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How fast can i run a mile with 3710 joules in my body?

Zinaida [17]

No conclusive evidence exists on “average” 1-mile run times, because there is no scientifically agreed-upon average runner. Opinion varies widely, but most anecdotal evidence places the average between seven and 10 minutes per mile for a non-competitive, in-shape runner.

7 0

3 years ago

a person brings his face close to a mirror . h e finds that the image of his face is magnified write the name of the mirror used

DIA [1.3K]

Answer:

Spherical concave mirrors

Explanation:

Like spherical convex mirrors, spherical concave mirrors have a focus. If the object is closer to the mirror than the focal point is, the image will be virtual, like we talked about before for the plane mirror and the convex mirror.

Concave mirrors, on the other hand, can have real images. If the object is further away from the mirror than the focal point, the image will be upside-down and real---meaning that the image appears on the same side of the mirror as the object.

The closer the object comes to the focal point (without passing it), the bigger the image will be.

You can try this yourself by looking into the concave side of a shiny spoon. If you look into the spoon while holding it at arm’s length, you’ll see an extremely magnified, upside-down image of your face. But as you bring the spoon closer to your eyes, the image will get bigger and bigger.

- Hope this helps! <3

5 0

3 years ago

Please answer correctlywill give the brainliestUrgent​

Please answer correctlywill give the brainliestUrgent