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pantera1 [17]
3 years ago
15

Explain how do winds cause surface currents?

Physics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer:

As wind or an ocean current moves, the Earth spins underneath it. ... The Coriolis effect bends the direction of surface currents to the right in the Northern Hemisphere and left in the Southern Hemisphere.

Explanation:

The Coriolis effect causes winds and currents to form circular patterns.

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I need help with my physics homework agh! Please help it's due tomorrow. <br>​
storchak [24]

Rubbing both pieces cause each piece to have a negative charge.

When two parts have the same they repel each other, so holding one piece up tot he end of the other piece would push it away.

Because one piece is held in the middle by a string, it would rotate the piece in a circle.

If they held the piece to the other end of the one held by a string it would start to rotate in the opposite direction.

4 0
3 years ago
Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien
otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
  • The height at which the block stops rising is approximately 1.1415 m
  • The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

  • The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

  • Length of the incline, x_{max} = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0
2 years ago
When ultra violets lights shine on glass what does it do to electrons in the glass structure?
andreev551 [17]

Answer:

No

Explanation:

7 0
2 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × \sqrt{\frac{m}{k} }   ................2

put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0
3 years ago
The amplitude of a paricular wave is 4.0 m. The crest to trough distance
kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0
2 years ago
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