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dlinn [17]
3 years ago
11

Ocean waves are traveling to the east at 3.2 m/s with a distance of 19 m between crests. (a) With what frequency do the waves hi

t the front of a boat when the boat is at anchor? Hz (b) With what frequency do the waves hit the front of a boat when the boat is moving westward at 1.5 m/s?
Physics
1 answer:
rjkz [21]3 years ago
8 0

Answer:

(a) Frequency will be 0.168 Hz

(b) Frequency will be 0.0789 Hz

Explanation:

(a) We have given velocity of the ocean waves in east direction v = 3.2 m /sec

Distance between the crest \Delta x=\lambda =19m

We know that v=\lambda f

So 3.2=19\times  f

f=0.168Hz

(b)Velocity of the ocean waves in westward direction v = 1.5 m /sec

Distance between the crest \Delta x=\lambda =19m

We know that v=\lambda f

So 1.5=19\times  f

f=0.0789Hz

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A car with an initial velocity of 5.0 is acelerated at 3.0for 4.0seconds what os the velocity of tje car ofter the 4.0seconds
sasho [114]
Vf =Vi+at
Vf= 5 + 3*4
Vf=5+12
Vf=17m/sec
6 0
3 years ago
A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied
Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

8 0
2 years ago
Compare and contrast microscopic and macroscopic energy transfer Give least three comparisons for each
Eduardwww [97]

Answer:

Differences

microscopic refers to substances visible to the naked eye

macroscopic are substances invisible to naked eye

Similarities

both refer to different scales that are useful to determining the size to different compounds.

Explanation:

Ima find more

3 0
3 years ago
A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove
jonny [76]

We must remember that the total net force equation at constant velocity is:

<span>F – Ff = 0</span>

of

F - µN = 0

Using Newton's 2nd Law of Motion:<span>

F = m a 

<span>Where,

F = net force acting on the body 
m = mass of the body 
a = acceleration of the body 

Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to 

F = net Force = 0 

Therefore, 

F - µN = 0 

where 

µ = coefficient of friction = 0.20 
N = normal force acting on the cart = 12 N 

Therefore, 

F - 0.20(12) = 0 

<span>F = 2.4 N </span></span></span>
4 0
3 years ago
Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th
Pie

Answer:

Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd

Explanation:

The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is

d = 12 + 9 = 21 yd

The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

d=\sqrt{12^2+9^2}=15 yd

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was

d = 15 - 3 = 12 yd

7 0
3 years ago
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