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skelet666 [1.2K]
3 years ago
11

Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second

s. The markings on the post submerged in water indicate that the water level fluctuates from a trough at 6.0 meters to a crest at 9.0 meters. What is the amplitude of the wave?
Physics
1 answer:
Anit [1.1K]3 years ago
3 0
The amplitude of a wave can be obtained by measuring the distance from the resting position of the wave to its crest. The resting position is half of the distance from the crest to the trough. Given that the distance between the crest and the trough is 3 meters, the amplitude should be half of that, which is 1.5 meters.
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Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re
Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

Learn more about Momentum here:

brainly.com/question/1042017

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4 0
1 year ago
Please I need help with these 2 questions. Thank you.
Lemur [1.5K]
First one is D and Second one is B
7 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
To calculate work done on an object, _____. A. multiply the force in the direction of motion by the distance the object moved B.
o-na [289]
I believe your answer would be B, hope it helps

6 0
3 years ago
Read 2 more answers
A student draws a picture of the products and reactants of a chemical reaction. What, if anything, is wrong with the drawing?
ASHA 777 [7]

The complete question is: A student draws a picture of the products and reactants of a chemical reaction. What, if anything, is wrong with the drawing?

A) The drawing is wrong because there are more chemicals on the products side.

B) The drawing is correct because there are 12 compounds on each side of the arrow.

C) The drawing is wrong because there are different compounds on each side of the arrow.

D) The drawing is correct because there are 12 atoms of each type on each side of the arrow.

Answer:

Option D is correct

Explanation:

In the diagram attached below, it can be seen that there are 12 atoms of element which combine with 12 atoms of another element forming a compound. For the drawing to be correct, there should be 12 atoms of each type of element on both the reactants as well as product side, which is the case. There cannot be imbalance in the number of atoms of different elements on the two sides for a chemical reaction to occur.

Hence, option D is correct.

5 0
3 years ago
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