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jolli1 [7]
2 years ago
13

A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

ater the last car passes the crossing with a speed of 15 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing. Answer in units of s.
Physics
1 answer:
hram777 [196]2 years ago
4 0

Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

v_i = 20.86 m/s

now the final speed of the train when it crossed the crossing

v_f = 15 km/h

v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

Now the time to cross that junction is given as

v_f - v_i = at

4.17 - 20.86 = a(-0.44)

t = 37.6 s

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A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to
dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence,i=30^{\circ}

Angle of refraction,r=19.24^{\circ}

a.Refractive index of air,n_1=1

We know that

n_2sinr=n_1sini

n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517

b.Wavelength of red light in vacuum,\lambda=632.8nm=632.8\times 10^{-9} m

1nm=10^{-9} m

Wavelength in the solution,\lambda'=\frac{\lambda}{n_2}

\lambda'=\frac{632.8}{1.517}=417nm

c.Frequency does not change .It remains same in vacuum and solution.

Frequency,\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}

Where c=3\times 10^8 m/s

Frequency,\nu=4.74\times 10^{14}Hz

d.Speed in the solution,v=\frac{c}{n_2}

v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s

5 0
3 years ago
Explain how the data collected and the calculations for the first and second resonance points in today's experiment would change
grandymaker [24]

Answer:

tssths

Explanation:

hgst

8 0
2 years ago
The metric unit to measure work which equals one newton meter<br> is called?
Solnce55 [7]
The metric unit to measure work which equals one newton meter is called One Joule.
8 0
3 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]

7 0
2 years ago
Two identical particles of charge 6 μμC and mass 3 μμg are initially at rest and held 3 cm apart. How fast will the particles mo
krek1111 [17]

Answer:

Explanation:

The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .

Their potential energy at distance d

= kq₁q₂ / d

= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J

= 16.2 J

Their total kinetic energy will be equal to this potential energy.

2 x 1/2 x mv² = 16.2

= 3 x 10⁻⁶ v² = 16.2

v = 5.4 x 10⁶

v = 2.32 x 10³ m/s

When masses are different , total P.E, will be divided between them as follows

K E of 3 μ = (16.2 / 30+3) x 30

= 14.73 J

1/2 X 3 X 10⁻⁶ v₁² = 14.73

v₁ = 3.13 x 10³

K E of 30 μ = (16.2 / 30+3) x 3

= 1.47 J

1/2 x 30 x 10⁻⁶ x v₂² = 1.47

v₂ = .313 x 10³ m/s

3 0
3 years ago
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