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Artyom0805 [142]
3 years ago
7

A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

ff refraction of the oil is 1.47, what is the angle of reflection?
Physics
1 answer:
Over [174]3 years ago
4 0

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

\theta_i = 34 degree

so we know that

\theta_i = \theta_r

so here we can say that reflection angle of light will be same as angle of incidence

\theta_r = 34 degree

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g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell
olga55 [171]

Answer:

The value is V  =  2 V

Explanation:

From the question we are told that

The length of the wire is l = 10 \ m

The current density is J =  4*10^6 \  A/m^2

The conductivity is \sigma  =  2*10^{7} \  S/m

Generally conductivity is mathematically represented as

\sigma  =  \frac{l}{RA}

Here R is the resistance which is mathematically represented as

R =  \frac{V}{I}

Here I is the current which is mathematically represented as

I  =  J * A

So

R =  \frac{V}{  J * A}

And

\sigma  =  \frac{l}{\frac{V}{  J * A} * A}

=> \sigma  =  \frac{l}{\frac{V}{J}}

=> V = \frac{l * J}{\sigma }

=> V = \frac{10  * 4*10^6}{2*10^{7}  }

=> V  =  2 V

5 0
3 years ago
What is the relationship between the strength of an
Anastaziya [24]

Answer:

As the number of turns in the coil increases, the strength  of the electromagnet increases.

Explanation:

When current flows through a coil the coil behaves as an electromagnet. The strength of electromagnet depend the amount of current, no of turns of coil and the core of coil.

B=μ₀ N I

μ₀ = permeability of the core

N = Number of turns of the coil

I = Current flowing through the coil

Increasing the current and number of coils increase the strength of electromagnet.

3 0
3 years ago
4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
KatRina [158]

Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

8 0
3 years ago
In an electronic transition an atom can not emit what?
iren [92.7K]
█ Question <span>█

</span><span>In an electronic transition, an atom cannot emit what?

</span>█ Answer █

When an electronic transition is occurring, an atom cannot emit ultra-violet light. 

<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>
7 0
3 years ago
An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration
lana [24]

Answer:

Vf= 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi= 10 m/s

Final velocity = Vf= 0 m/s

Gravity acting downward =  g = -9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²

( Acceleration is consider negative if it is in opposite direction of velocity )

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) *s = 0 - 10²

⇒-25.6 *s = -100

⇒ s = 100/ 25.6

⇒ s = 3.9 m

Downward motion:

Vi= 0 m/s

s = 3.9 m

Gravity acting downward =  g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁  =  9.8 - 3  = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf= 7.29 m/s

8 0
3 years ago
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