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True [87]
2 years ago
5

A car is being driven down a level track with no net force on it. This motion is best described as

Physics
1 answer:
Alex2 years ago
3 0

A)Linear motion

If there is not net force on the car, then by the Newton Second Law, the acceleration is zero, and the only valid option for zero acceleration is A).

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A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra
lidiya [134]

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin  r

 r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

tan \alpha = \dfrac{t}{d/2}

tan 61^0 = \dfrac{t}{2.5/2}

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

4 0
3 years ago
A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has
Zepler [3.9K]

The answer is the second option, or 1/10 the same momentum.

7 0
2 years ago
Read 2 more answers
A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas
Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

(c) the maximum acceleration of the object, a_max = 15.4 m/s²

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

V_{max} =\omega*A=  \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s

(c) the maximum acceleration of the object

a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2

6 0
2 years ago
Parasaurolophus was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.3 m long hollow tube in the cr
yarga [219]

Answer:

67.31 Hz, 201.92 Hz, 336.54 Hz

Explanation:

Resonance also known as standing wave occur when a uniform tube is closed at one end and open at the other end.  

For a resonance wave in a tube with length b, wavelength λ of resonance wave is 4b because a quarter wave is formed in the tube, that is,  

           λ = 4b.  

When the speed of sound is c, the frequency of sound, f, is  

           f = c /λ,  

and therefore,  

           f = c / ( 4b ).  

From the question,

When c = 350 m/s and b = 1.3 m,  

using the formula            

fn = c / ( 4b ) × n (n = 1, 3, 5, ...).  

The first three resonant frequencies are;

First resonance frequency is

f = 350 / (4*1.3) * 1 = 67.31 Hz

Second resonance frequency is

f = 350 / (4*1.3) * 3 = 201.92 Hz

Third resonance frequency is

f = 350 / (4*1.3) * 5 = 336.54 Hz

5 0
3 years ago
How does the current change if the resistance is doubled?
finlep [7]
The current will be divided by 2
7 0
3 years ago
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