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andrezito [222]
3 years ago
14

PLS HELP Do you believe this relationship between incident and reflected angles would occur even if the medium interface were cu

rved, like a curved mirror? Justify your response.
Physics
1 answer:
Anna35 [415]3 years ago
4 0
The Law of reflection would still hold even off a curved surface. Since the angles are measured from the normal, which is perpendicular to the surface, curved surfaces don't matter. This is basis of curved mirrors such as concave and convex 
You might be interested in
A stream moving with a speed of 7.1 m/s reaches a point where the cross-sectional area of the stream decreases to one half of th
lana66690 [7]

Answer:

14.2 m/s

Explanation:

Given data:

Speed of the stream, v₁ = 7.1 m/s

let the cross section area at initial point be A₁

now area at the second point, A₂ = (1/2)A₁ = 0.5A₁

now, from the continuity equation, we have

A₁v₁ = A₂v₂

where, v₂ is the velocity at the narrowed portion

thus, on substituting the values, we get

A₁ × 7.1 = 0.5A₁ × v₂

or

v₂ = 14.2 m/s

8 0
3 years ago
Armand hands the rock to Pierre, who is standing next to the trampoline. Explain how Armand moves, in terms of the forces acting
Nikolay [14]

Answer:

Armando's weight ,restored force created by the trampoline

a harmonic movement within the trampoline

Explanation:

In a trampoline we have two forces that actuate Armando's weight and the restored force created by the trampoline that depends on the deformation distance of the elastic canvas.

Amando's weight is vertical and directed towards the center of the Earth and has a constant value, this weight is balanced with the elastic force the springboard exerts on Armando in a vertical direction.

In general, when entering the trampoline, a small jump is made, this creates a speed that deforms the canvas until the speed is reduced to zero, at this point the elastic force is greater than the weight and the boy begins to climb, After the boy leaves the canvas he meets only the force of gravity and his speed decreases to zero and begins his fall.

In Summary Armando is in a harmonic movement within the trampoline

6 0
3 years ago
How do you do this question? Please include free body diagrams and clear explanation, so I can understand.
vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.

∑F = ma

86.5 N − T₁ − Wa = 0

Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

ma = 3.2 kg

Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

mb × 9.8 m/s² = 55.6 N − 36.5 N

mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

mc × 9.8 m/s² = 36.5 N − 9.6 N

mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

3 0
3 years ago
A circular ring with area 4.45 cm2 is carrying a current of 13.5 A. The ring, initially at rest, is immersed in a region of unif
Gwar [14]

Answer:

a) ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ ) N.m

b) ΔU = -0.000747871 J

c)  w = 47.97 rad / s

Explanation:

Given:-

- The area of the circular ring, A = 4.45 cm^2

- The current carried by circular ring, I = 13.5 Amps

- The magnetic field strength, vec ( B ) = (1.05×10−2T).(12i^+3j^−4k^)

- The magnetic moment initial orientation, vec ( μi ) = μ.(−0.8i^+0.6j^)  

- The magnetic moment final orientation, vec ( μf ) = -μ k^

- The inertia of ring, T = 6.50×10^−7 kg⋅m2

Solution:-

- First we will determine the magnitude of magnetic moment ( μ ) from the following relation:

                    μ = N*I*A

Where,

           N: The number of turns

           I : Current in coil

           A: the cross sectional area of coil

- Use the given values and determine the magnitude ( μ ) for a single coil i.e ( N = 1 ):

                    μ = 1*( 13.5 ) * ( 4.45 / 100^2 )

                    μ = 0.0060075 A-m^2

- From definition the torque on the ring is the determined from cross product of the magnetic moment vec ( μ ) and magnetic field strength vec ( B ). The torque on the ring in initial position:

             vec ( τi ) = vec ( μi ) x vec ( B )

              = 0.0060075*( -0.8 i^ + 0.6 j^ ) x 0.0105*( 12 i^ + 3 j^ -4 k^ )

              = ( -0.004806 i^ + 0.0036045 j^ ) x ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

- Perform cross product:

          \left[\begin{array}{ccc}i&j&k\\-0.004806&0.0036045&0\\0.126&0.0315&-0.042\end{array}\right]  = \left[\begin{array}{ccc}-0.00015139\\-0.00020185\\-0.00060556\end{array}\right] \\\\

- The initial torque ( τi ) is written as follows:

           vec ( τi ) = ( 0.0015139 i^ + 0.0020185 j^ + 0.00060556 k^ )

           

- The magnetic potential energy ( U ) is the dot product of magnetic moment vec ( μ ) and magnetic field strength vec ( B ):

- The initial potential energy stored in the circular ring ( Ui ) is:

          Ui = - vec ( μi ) . vec ( B )

          Ui =- ( -0.004806 i^ + 0.0036045 j^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Ui = -[( -0.004806*0.126 ) + ( 0.0036045*0.0315 ) + ( 0*-0.042 )]

          Ui = - [(-0.000605556 + 0.00011)]

          Ui = 0.000495556 J

- The final potential energy stored in the circular ring ( Uf ) is determined in the similar manner after the ring is rotated by 90 degrees with a new magnetic moment orientation ( μf ) :

          Uf = - vec ( μf ) . vec ( B )

          Uf = - ( -0.0060075 k^ ) . ( 0.126 i^ + 0.0315 j^ -0.042 k^ )

          Uf = - [( 0*0.126 ) + ( 0*0.0315 ) + ( -0.0060075*-0.042 ) ]

          Uf = -0.000252315 J

- The decrease in magnetic potential energy of the ring is arithmetically determined:

          ΔU = Uf - Ui

          ΔU = -0.000252315 - 0.000495556  

          ΔU = -0.000747871 J

Answer: There was a decrease of ΔU = -0.000747871 J of potential energy stored in the ring.

- We will consider the system to be isolated from any fictitious forces and gravitational effects are negligible on the current carrying ring.

- The conservation of magnetic potential ( U ) energy in the form of Kinetic energy ( Ek ) is valid for the given application:

                Ui + Eki = Uf + Ekf

Where,

             Eki : The initial kinetic energy ( initially at rest ) = 0

             Ekf : The final kinetic energy at second position

- The loss in potential energy stored is due to the conversion of potential energy into rotational kinetic energy of current carrying ring.    

               -ΔU = Ekf

                0.5*T*w^2 = -ΔU

                w^2 = -ΔU*2 / T

Where,

                w: The angular speed at second position

               w = √(0.000747871*2 / 6.50×10^−7)

              w = 47.97 rad / s

6 0
3 years ago
If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?
nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = \frac{GM}{R^{2} }

where G is the gravitational constant

Gravitational force of the planet = \frac{G4M}{(4R)^{2} }

                                                       =\frac{G4M}{16R^{2} }

                                                       =\frac{GM}{4R^{2} }

recall, gravitational force of earth is given as = \frac{GM}{R^{2} }

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0
3 years ago
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