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3 years ago

If you are in a car that is being pulled down a 56 m path with a force of 12.5

Physics

1 answer:

Elis [28]3 years ago

7 0

Answer:

answer is 700 joules

Explanation:

work is equal to force into displacement

W=12.5*56

700 joules

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A ray of light (f = 5.09 x 1014 Hz) is incident on the boundary between air and an unknown material X at an angle of incidence o

11Alexandr11 [23.1K]

Answer:

A) Material X is flint glass

B) Speed of light in the X material is 1.807 x $10^{8}$ m/s

C) Angle of refraction of light in medium X is $29.57^{o}$

Explanation:

Given data:

frequency of the light f = 5.09 x $10^{14}$ Hz

angle of incidence Θ $_{i}$ = $55^{o}$

index of refraction of material $n_{2}$ = 1.66

A) To find material X

Given the index of refraction is 1.66 and hence the material is the flint glass

B) To calculate the speed of light in the material.

We know that the relation between index of refraction (n), velocity of light (c = 3 x $10^{8}$ m/s) and velocity of light is given by the equation:

n = c/v

Hence,

Speed of light in the X material v = c/n

= $\frac{3 x 10^{8} }{1.66}$

= 1.807 x $10^{8}$ m/s

C) To calculate angle of refraction of light in medium X

We know that the Snell's law states that

$n_{1} sin$ Θ $_{i}$ = $n_{2} sin$ Θ $_{r}$

$n_{1}$ = incident index

$n_{2}$ = refracted index

Θ $_{i}$ = incident angle

Θ $_{r}$ = refracted angle

In given problem, $n_{1}$ = 1 since medium is air

Substituting the known values, we get

1 x $sin$ $55^{o}$ = 1.66 x $sin$ Θ $_{r}$

$sin$ Θ $_{r}$ = $sin$ $55^{o}$ /1.66

= 0.4935

Hence, angle of refraction of light in medium X Θ $_{r}$ = $sin^{-1}$ (0.4935)

= $29.57^{o}$

4 0

3 years ago

What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w

kipiarov [429]

156m\s
Hope this helps! :)

5 0

3 years ago

Read 2 more answers

How high is a 3 kg object that has 3000 j of potential energy

Mila [183]

Using the formula $z=\frac{E}{m*g}$

$z=\frac{3000}{3*9.80665}$

$z=101.97$ meters

8 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant

Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

$emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}$ $\ \ =-B\frac{A_2-A_1}{t_2-t_1}$ (1)

ФB: magnetic flux = BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

$c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm$

To find the areas A1 and A2 you calculate the radius:

$r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm$

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

$A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2$

Finally, the emf induced, by using the equation (1), is:

$emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV$

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0

3 years ago

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

Marta_Voda [28]

The net force acting on the object perpendicular to the table is

∑ F[perp] = F[normal] - mg = 0

where mg is the weight of the object. Then

F[normal] = mg = (15 kg) (9.8 m/s²) = 147 N

The maximum magnitude of static friction is then

0.40 F[normal] = 58.8 N

which means the applied 40 N force is not enough to make the object start to move. So the object has zero acceleration and does not move.

8 0

2 years ago