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Fantom [35]
3 years ago
11

A buoy is anchored to the ocean floor a large wave approaches the buoy how will the buoy move as the wave goes by

Physics
1 answer:
I am Lyosha [343]3 years ago
7 0
Alvin (DSV-2) is a manned deep-ocean research submersible owned by the United States Navy and operated by the Woods Hole Oceanographic Institution (WHOI) in Woods Hole, Massachusetts. The vehicle was built by General Mills' Electronics Group[2] in Minneapolis, Minnesota. Named to honor the prime mover and creative inspiration for the vehicle, Allyn Vine, Alvin was commissioned on 5 June 1964. The submersible is launched from the deep submergence support vessel RV Atlantis (AGOR-25), which is also owned by the U.S. Navy and operated by WHOI. The submersible has made more than 4,400 dives, carrying two scientists and a pilot, to observe the lifeforms that must cope with super-pressures
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Picture attached please help! 2 different questions!
Maksim231197 [3]

Wave A would have higher amplitude

Hope this helps :D

7 0
3 years ago
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What is the force exerted on a moving charge of –2.0 μC at a 20° angle through a magnetic field of 3.0 × 10–4 T with a velocity
san4es73 [151]
1.0 x 10-3 N is the correct answer!
6 0
3 years ago
Read 2 more answers
Please help Math Phys
Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land.  Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0
3 years ago
Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic
VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

m_Agh=\frac{1}{2}m_Av_A^2

where

g=9.8 m/s^2 is the acceleration due to gravity

m_A=0.5 kg is the mass of the block

v_A is the speed of the block A just before touching block B

Solving for the speed,

v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

e=\frac{v'_B-v'_A}{v_A-v_B}

where:

e = 0.7 is the coefficient of restitution in this case

v_B' is the final velocity of block B

v_A' is the final velocity of block A

v_A=3.83 m/s

v_B=0 is the initial velocity of block B

Solving,

v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s

Re-arranging it,

v_A'=v_B'-2.68 (1)

Also, the total momentum must be conserved, so we can write:

m_A v_A + m_B v_B = m_A v'_A + m_B v'_B

where

m_B=2 kg

And substituting (1) and all the other values,

m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m

5 0
3 years ago
A uniform bar 12 m long has a mass 20 kg. A 10 kg pointed mass is attached to one end of the rod. Now the position of COM of the
user100 [1]

Answer:

.

Explanation:

/

7 0
3 years ago
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