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Georgia [21]
3 years ago
13

The angle of reflection is the angle the ____ to the reflecting surface.

Physics
1 answer:
Arlecino [84]3 years ago
7 0

Answer:

The angle of reflection is the angle the reflected rays make with a perpendicular line to the reflecting surface.  

Explanation:

Reflection It is the change of direction suffered by a luminous ray when hitting the surface of an object. The angle of reflection is that which is formed by the reflected ray and the normal vector to the study surface

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A rock rolls down a hill. Which form of energy is this an example of? (2 points)
Bond [772]

Answer:

c.Mechanical

Explanation:

6 0
3 years ago
You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim
andreev551 [17]

Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

distance traveled in the reaction time d = v t

d = 21 * 0.50 = 10.5 m

deceleration after this time = -10 m/s^2

now the distance traveled by the car after applying bakes

v_f^2 - v_i^2 = 2a d

0 - 21^2 = 2(-10)d

d = 22.05 m

so total distance moved before it stop

d = 22.05 + 10.5 = 32.55 m

so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

so our distance after brakes are applied is d = 35 - 10.5 = 24.5 m

again by kinematics

v_f^2 - v_i^2 = 2 ad

0 - v^2 = 2(-10)(24.5)

v = 22.1 m/s

so maximum speed would be 22.1 m/s

5 0
3 years ago
Se um corpo 1 está em movimento em relação a um corpo 2, e se o corpo 2 está em repouso em relação a um corpo 3, pode-se afirmar
stealth61 [152]
Yo habla ingles no espanol
8 0
3 years ago
Read 2 more answers
A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec
AnnyKZ [126]
2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.
5 0
3 years ago
An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
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