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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 
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Is this good? Its for a science test ~
Digiron [165]
Yes..............................
4 0
3 years ago
A 326 g object is attached to a spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the
Basile [38]

Answer:

a) v_{max}=5.98\frac{m}{s}

b) k=205.92\frac{N}{m}

c) A=0.24m

Explanation:

a) In the equilibrium position of the system, that is when the spring is not elongated, the potential energy is zero. Therefore, the total energy of the system is the maximum kinetic energy:

E=K_{max}\\E=\frac{mv_{max}^2}{2}\\v_{max}=\sqrt{\frac{2E}{m}}\\v_{max}=\sqrt{\frac{2(5.83J)}{0.326kg}}\\v_{max}=5.98\frac{m}{s}

b) The force constant of the spring can be calculated from the natural frequency of the system:

\omega^2=\frac{k}{m}\\k=m\omega^2

Recall that \omega=\frac{2\pi}{T}\\, that is the distance traveled in one revolution divided into the time of one revolution. Replacing and solving for k:

k=\frac{m4\pi^2}{T^2}\\k=\frac{(0.326kg)4\pi^2}{(0.25s)^2}\\k=205.92\frac{N}{m}

c) The maximum speed is directly proportional to the amplitude of the motion:

v_{max}=A\omega\\A=\frac{v_{max}}{\omega}\\A=\frac{(v_{max})T}{2\pi}\\A=\frac{(5.98\frac{m}{s})0.25s}{2\pi}\\A=0.24m

3 0
3 years ago
What is the mass of a crane that has a weight of 697,005.40N?
igomit [66]

Explanation:

hope this helps you out if not im sorry

6 0
2 years ago
An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9
aleksandr82 [10.1K]

Answer:

The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²

The rocket's motion for analysis sake is divided into two phases.

Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m

Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.

Explanation:

The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.

The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.

The detailed step by step solution to the problems can be found in the attachment below.

Thank you and I hope this solution is helpful to you. Good luck.

5 0
3 years ago
A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and
N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t    where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec =  1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r  in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0
3 years ago
Read 2 more answers
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