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2 years ago

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in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m

ass. If the tire marks show that the two vehicles moved after the collision at 4 m/s, what was the speed of the vehicle before the collision

1 answer:

baherus [9]2 years ago

6 0

The initial speed of the vehicle before the collision is 8 m/s.

Let the mass of the vehicle = m
Let the initial speed of the vehicle stopped = u
The initial speed of the vehicle parked at the red light = 0

<h3>Principle of conservation of linear momentum</h3>

The initial speed of the vehicle before the collision is calculated by applying principle of conservation of linear momentum as follows;

$m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s$

Thus, the initial speed of the vehicle before the collision is 8 m/s.

Learn more about inelastic collision here: brainly.com/question/7694106

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Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi

Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

$\delta = m\lambda$ (1)

where

$\delta =|d_1 -d_2|$ is the path difference, with

$d_1$ is the distance of the point from the first source

$d_2$ is the distance of the point from the second source

m is an integer number

$\lambda$ is the wavelength

In this problem, we have

$d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m$

So let's use eq.(1) to see if the resulting m is an integer

$\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5$

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

$\delta = (m+\frac{1}{2})\lambda$ (2)

If we apply the same procedure to eq.(2), we find

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2$

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

$d_1 = 91.0 m\\d_2 =221.0 m$

So the path difference is

$\delta =|91.0 m-221.0 m|=130.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5$

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

$d_1 = 44.0 m\\d_2 =135.0 m$

So the path difference is

$\delta =|44.0 m-135.0 m|=91.0 m$

Using the condition for constructive interference:

$m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5$

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

$m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3$

which is an integer: so, this point is a point of destructive interference.

3 0

3 years ago

When a beam of charged particles moves through a magnetic field, what is the evidence that particles in the beam have momentum g

Vsevolod [243]

The charged particles are often deflated in a magnetic field.

<h3>What is a magnetic field?</h3>

The term charge refers to a positive or negative entity. The can be created when a charge is made to pass through a conductor in a magnetic field.

A magnetic field is created when we have a north pole and a south pole. The charged particles could be made to pass through the electric field and when that happens, we can see a pattern a shown in the image attached.

Thus, we can see that the charged particles are often deflated in a magnetic field.

Learn more about magnetic field:brainly.com/question/23096032

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4 0

1 year ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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3 years ago

Which pair of equations can describe the path of a particle moving with an acceleration that is perpendicular to the velocity of

antoniya [11.8K]

Answer:

The question clearly describes the circular motion.

The circular motion equation is

$a_{radial} = \frac{v^2}{r}$

The path of the particle is circular.

Explanation:

In circular motion, the radial acceleration is always towards the center and constant in magnitude. Furthermore, the velocity of the circular motion is always tangential to the circle, that is it is always perpendicular to the radius, hence the acceleration.

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2 years ago

If the pressure in a gas is doubled while its volume is held constant, by what factor do vrms change

Nat2105 [25]

Answer is given below

Explanation:

given data

pressure = double

volume = constant

solution

As we know that an Average velocity and rms velocity is directly proportional to square root of PV ..................1

so if we take P is doubled while keeping V constant

than Velocity increases by a factor $\sqrt{2}$

so that Factor = 1.414 for both the cases

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3 years ago