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levacccp [35]
4 years ago
5

((PLEASE HELP))

Physics
2 answers:
Degger [83]4 years ago
5 0

Answer:

Angle of refraction

Explanation:

The incident ray is the ray before it reaches the surface.

The refracted ray is the ray after it reaches the surface.

n₁ is called the index of incidence.

n₂ is called the index of refraction.

θ₁ is called the angle of incidence.

θ₂ is called the angle of refraction.

They are related by Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

il63 [147K]4 years ago
5 0

Answer:

Angle of refraction

Explanation:

The angle θ2 represents the angle of refraction.

Refraction is the change in direction of light rays as it passes from one medium to another. As light rays passes from one medium to another, it changes direction.

If the incident light is passing from the less dense medium to denser medium, the refracted ray will bend towards the normal. In this case the angle of refraction 'r' is always less than the angle if incidence. (According to the diagram)

Similarly, If the incident light is passing from the denser medium to less dense medium, the refracted ray will bend away from the normal. In this case the angle of refraction 'r' is always greater than the angle of incidence.

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Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138
ira [324]

Answer: 276 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:  

A=A_{o}.2^{\frac{-t}{h}} (1)  

Where:  

A=\frac{1}{4}A_{o} is the final amount of the material

A_{o} is the initial amount of the material  

t is the time elapsed  

h=138 days is the half life of polonium-210

Knowing this, let's substitute the values and find t from (1):

\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}} (2)  

\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}} (3)  

\frac{1}{4}=2^{\frac{-t}{138 days}} (4)  

Applying natural logarithm in both sides:

ln(\frac{1}{4})=ln(2^{\frac{-t}{138 days}}) (5)  

-1.386=-\frac{t}{138days}ln(2) (6)  

Clearing t:

t=276days (7)  

7 0
3 years ago
A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, w
Masteriza [31]

Answer:U(x) = 30x^2 +6x^3

V^2=8.28m/s

Explanation:The law of conservation of energy is given by K1+U1= K2+U2 ...eq 1

Kinetic energy K.E= 1/2 mv^2

Restoring force function F(x)= -60x - 18x^2

But F(x)= -dU/dx

dU(x)=-F(x)dx

Integrating U(x)= -integral F(x)dx + U(0)

Substituting, we get

U(x) = - integral(-60x-18x^2)dx+U(0)

U(x)= 30x^2+6x^3+U(0)

U=0 at x=0

Therefore U(x)= 30x^2+6x^3

b) Given : x1=1.00m,x2= 0.50m ,V1=0, V2=?

Substituting into eq (a)

U1= 30(1.00)^2+6(1.00)^3=36J

Using x2=0.5 into eq(a)

U2=30(0.50)^2+6(0.50)^3=8.25J

Object at rest K1=0

0+36=K2+8.25

K2=27.75J

Given; m =0.900kg, V2=?

27.75=1/2×0.900×V2^2

V2= SQRT(2×27.75)/0.81

V2= 8.28m/s

4 0
4 years ago
Two spheres A and B of negligible dimensions and masses 1 kg and √3 kg respectively, are supported on the smooth circular surfac
alexdok [17]

Answer:

α = π/3

β = π/6

Explanation:

Use arc length equation to find the sum of the angles.

s = rθ

π/20 m = (0.1 m) (α + β)

π/2 = α + β

Draw a free body diagram for each sphere.  Both spheres have three forces acting on them:

Weight force mg pulling down,

Normal force N pushing perpendicular to the surface,

and tension force T pulling tangential to the surface.

Sum of forces on A in the tangential direction:

∑F = ma

T − m₁g sin α = 0

T = m₁g sin α

Sum of forces on B in the tangential direction:

∑F = ma

T − m₂g sin β = 0

T = m₂g sin β

Substituting:

m₁g sin α = m₂g sin β

m₁ sin α = m₂ sin β

(1 kg) sin α = (√3 kg) sin (π/2 − α)

1 sin α = √3 cos α

tan α = √3

α = π/3

β = π/6

4 0
4 years ago
A 1.3 kg booster is attached to a 6.0 kg rocket is initially travelling at a velocity of 175 m/s. If the booster is ejected in b
nikklg [1K]

Answer:

It will travel to the infinity and beyond

4 0
4 years ago
A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
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