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3 years ago

Sound cannot travel through a vacuum. Explain why.

Physics

1 answer:

asambeis [7]3 years ago

3 0

Answer:

A vacuum is an area without any air, like space. So sound cannot travel through space because there is no matter for the vibrations to work on it.

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What are the disadvantages of driverless cars? Check all that apply

Sauron [17]

Answer:

Explanation:

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2 years ago

A motorist enters a freeway at 45 km/h and accelerates uniformly to 99 km/h. From the odometer in the car, the motorist knows th

Marat540 [252]

Answer:

a) 19440 km/h²

b) 10 sec

Explanation:

v₀ = initial velocity of the car = 45 km/h

v = final velocity achieved by the car = 99 km/h

d = distance traveled by the car while accelerating = 0.2 km

a = acceleration of the car

Using the kinematics equation

v² = v₀² + 2 a d

99² = 45² + 2 a (0.2)

a = 19440 km/h²

t = time required to reach the final velocity

Using the kinematics equation

v = v₀ + a t

99 = 45 + (19440) t

t = 0.00278 h

t = 0.00278 x 3600 sec

t = 10 sec

5 0

2 years ago

A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same mome

Kaylis [27]

a) since force = mass * acceleration

f= 900 * 0 (because constant speed = 0 acceleration)

similarly b) f = 0

7 0

3 years ago

Read 2 more answers

Who was the first person to say that earth orbits the sun?

Whitepunk [10]

The first person to say the Earth orbited the sun was Nicolaus Copernicus

7 0

3 years ago

Read 2 more answers

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

The mass is $m_2 =21.75*10^{-27} \ kg$

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago