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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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Find the dimension of volume​
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Answer:

L^3

Explanation:

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The difference in energy on either side of the wheel is a measure of the wheel’s ___________.
atroni [7]
I believe the answer is potential difference
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A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

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k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0
3 years ago
Read 2 more answers
At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A
saveliy_v [14]

Answer with Explanation:

We are given that

Magnetic field,B=0.6\times 10^{-4} T

\theta=75^{\circ}

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F=IBlsin\theta

Using the formula

F=0.6\times 10^{-4}\times 15\times 19sin75

F=16.5\times 10^{-3} N

Direction=tan\theta=cot(90-75)=tan15^{\circ}

\theta=15^{\circ}

15 degree above the horizontal  in the northward direction.

5 0
3 years ago
A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

dy = \sqrt{d^2 - dx^2}

dy = \sqrt{3.85^2 - 3^2}  = 2.41 km

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0
3 years ago
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