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nikdorinn [45]
4 years ago
15

A wave is incident on the surface of a mirror at an angle of 41° with the normal. What can you say about its angle of reflection

?
Physics
2 answers:
bazaltina [42]4 years ago
7 0
Hold on 63 deg angle my friend
LenaWriter [7]4 years ago
3 0
A wave incident on the surface of a mirror at an angle of 41 degrees with the normal. What can you say about it's angle of reflection? 
You can say angle of reflection angle of incidence is always equivalent to angel of reflection.
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A 20 kg wombat climbs 10 m up a tree. When it stops what is the wombats gravitational potential energy?
swat32

Answer:

2000J

Explanation:

PE = m*g*h

PE= 20kg*10m/s^-2*10m

PE= 2000J

5 0
4 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Anon25 [30]

In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy.  That idea is easy to apply to this question.

8 0
4 years ago
Read 2 more answers
Find the vector representing the area of the rectangle with vertices and oriented so that it faces downward. The magnitude of th
sergejj [24]

Answer:

The answer is  "-72 k".

Explanation:

Please find the complete question in the attached file.

Given point:

A=(0,0,0)\\B=(0,8,0)\\C=(9,8,0)\\D=(9,0,0)

\bar{AB} = (0i+8j+0k)-(0i+0j+0k)= 8j\\\\\bar{AC} = (9i+8j+0k)-(0i+0j+0k)= 9i+8j\\\\

Calculating the area:

Area=\left|\begin{array}{ccc}i&j&k\\0&8&0\\9&8&0\end{array}\right|

       =i[8(0)-8(0)]-j[(0-0)]+k[(0-9(8))]\\\\=i[0-0]-j[(0)]+k[(0-72)]\\\\=i[0]-j[(0)]+k[(-72)]\\\\=-72 \ k

3 0
3 years ago
A racing car whose mass is 1.2 X 10^3 kg is travelling at 8.9 m/s. It stops with a constant deceleration in a distance of 1.8X10
Alexeev081 [22]

given that initial speed of the car is

v_i = 8.9 m/s

now after travelling the distance d = 1.8 * 10^1 m the car will stop

so here we can use kinematics to find the acceleration of car

v_f^2 - v_i^2 = 2 a d

0 - 8.9^2 = 2 a d

here we have

- 79.21 = 2*(18)*a

a = -2.2 m/s^2

net force applied due to brakes of car is given by Newton's II law

F = ma

here we have

mass = 1.2 * 10^3 kg

F_{net} = 1.2 * 10^3 * 2.2

F_{net} = 2.64 * 10^3 N

now we can say

F_{net} = F_1 + F_2

2.64 * 10^3 = 1.8 * 10^3 + F_2

F_2 = 8.4 * 10^2 N

So the force applied due to brakes is given as above

7 0
3 years ago
Hat do peaked roofs, convertible tops, and airplane wings have in common when air moves faster across their top surfaces? What d
Nutka1998 [239]

Answer:

a.

Explanation:

The air moves faster across the top of the surface and exerts less pressure on the air below and thus due to difference in air pressure the top roof is lifted.

hence the correct option is pressure underneath them is reduced.

6 0
3 years ago
Read 2 more answers
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