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3 years ago

An object accelerates from rest, with a constant acceleration of 14.5 m / s, what will its velocity be after 92 s?

Physics

1 answer:

aniked [119]3 years ago

6 0

Answer:

If an object accelerates from rest, with a constant acceleration of 5.4 m/s2, what will its velocity be after 28s? v0 = 0 v= ? a = 5.4 m/s2. ∆x = t = 28 s. 3.

Explanation:

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What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertzDoes sound travel faster in a

babymother [125]

Answer:

Wavelength of wave is 2.5 meters

Explanation:

Given that,

Speed of a wave, v = 350 m/s

Frequency of the wave, f = 140 Hz

The formula to find the speed of sound is given by :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{350\ m/s}{140\ Hz}\\\\\lambda=2.5 m$

The speed of sound is more in higher temperature. At higher temperature, the molecules vibrate with more energy. As a result, sound travel faster in a warm room.

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3 years ago

A hilly road is spiral in shape.

Llana [10]

I- wym incorrect answer i-

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3 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ (we use this approximation instead of the original $d^2+a^2\approx a^2$ since $z$ , our assumption still holds )

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

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4 years ago

What is the difference between a steep and shallow slope

Airida [17]

Whether a line is "steep" or "shallow" depends on the slope of the line. In the equation y=mx+b, m is the slope. In your first equation, 1/2x+4, the slope would be 1/2 which means there is 1 increase in the vertical, or y, value for each 2 increases in the horizontal, or x value. This line would be shallow.

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4 years ago

Read 2 more answers

if i release one steel ball from the top of a ramp and the other ball from the 40cm mark will they have the same acceleration?

solniwko [45]

Answer:

Yes, they will also have the same acceleration. Acceleration is controlled by the amount of weight (def. amount of gravitational pull on a given object) that the ball has. For one ball to accelerate quicker than the other, it would need a propulsion element, which it does not.

Explanation:

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2 years ago