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viva [34]
2 years ago
3

A Ferrari with a mass of 1400 kg approaches a freeway underpass that is 12 m across. Part A At what speed must the car be moving

, in order for it to have a wavelength such that it might somehow "diffract" after passing through this "single slit"?
How do these conditions compare to normal freeway speeds of 30 ms?
Physics
1 answer:
Shalnov [3]2 years ago
7 0

To solve this problem, we will apply the concepts related to momentum both for an undulating particle given by the Planck constant and the wavelength, as well as the product between the mass and the speed. We will match the terms and then compare

\text{Mass of the Ferrari} = m = 1400kg

\text{The Wavelength} = \lambda = 10m

The momentum of the car is given as,

P = \frac{h}{\lambda}

Here,

h = \text{Plank's constant} = 6.63*10^{-34}J\cdot s

The momentum of the car related with the speed is

P = mv

Equation we have

mv = \frac{h}{\lambda}

v = \frac{h}{m\lambda}

Replacing with our values we have that,

v = \frac{(6.63*10^{-34}J\cdot s)}{(1400kg)(12m)}

v = 3.946*10^{-38}m/s

Now the wavelength will be

\lambda = \frac{h}{mv}

\lambda = \frac{(6.63*10^{-34}J\cdot s)}{(1400kg)(30m/s)}

\lambda = 1.578*10^{-38}m << 12 m

Therefore the speed is 30m/s, then we get the wavelength is less than 12m. The diffraction is not occurring

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Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to
Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate m=100 kg

Crate slows down in s=1.5 m

initial speed u=1.77 m/s

inclination \theta =30^{\circ}

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2

W_{gravity}=mg(0-h)=mgs\sin \theta

W_{gravity}=-mgs\sin \theta

W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N

change in kinetic energy=\frac{1}{2}\times 100\times 1.77^2=156.64 J

W_{friction}=156.64+735=891.645

(b)Coefficient of sliding friction

f_r\cdot s=W_{friciton}

891.645=f_r\times 1.5

f_r=594.43 N

and f_r=\mu mg\cos \theta

\mu 100\times 9.8\times \cos 30=594.43

\mu =0.7

5 0
2 years ago
A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polar
loris [4]

Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

For the second polarizer, the intensity is equal:

I_{2} =\frac{I_{o} }{2} (cos\theta )^{2} =\frac{I_{o} }{2} (cos90)^{2} =0

5 0
2 years ago
You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly
Viefleur [7K]

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

7 0
2 years ago
Read 2 more answers
A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh
Greeley [361]

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

4 0
2 years ago
Read 2 more answers
A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met
faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0
2 years ago
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