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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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What will happen when the sun blows up?
RSB [31]
We will all chaotically burn to death!



Hope this helps!!

8 0
3 years ago
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A load of 54 N attached to a spring that ishanging vertically stretches the spring 0.15 m.What is the spring constant?Answer in
beks73 [17]

Answer:

300 N/m

Explanation:

given,

Load attached to the spring, W = 54 N

length of stretch of the spring, x = 0.15 m

spring constant= ?

Force applied on the spring is calculated by the equation

F = k x

where k is the spring constant

x is the displacement of the spring due to applied load

now,

54 = k × 0.15

k = \dfrac{54}{0.15}

k =300\ N/m

hence, the spring constant is equal to 300 N/m

8 0
3 years ago
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"
Volgvan

Answer:

The height is : 60.025 m

Explanation:

The flowerpot falls off the balcony with zero launch angle

Given the time of fright as 3.5 s then ;

The formula to apply is ;

T=\sqrt{\frac{2H}{g} }\\\\3.5=\sqrt{\frac{2H}{9.8} }

3.5²= 2H/9.8

12.25 =2H/9.8

12.25 * 9.8 = 2H

120.05 = 2H

120.05/2 = H

60.025 =H

7 0
3 years ago
Choose the answer to fill in the blank, which best completes the sentence.
Allisa [31]

Answer:

A on edge 2020

Explanation:

4 0
3 years ago
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After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T
motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

KE_r = \dfrac{1}{2}I\omega^2

I of the moment of inertia of the sphere

I = \dfrac{2}{5}MR^2

 v = R ω

using conservation of energy

KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2

KE_r = \dfrac{1}{5}MV^2

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h

0.7 V^2 = 0.7 V'^2 + gh

0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0
3 years ago
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