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3 years ago

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Which of these is an example of a mechanical wave? A. the light of the Sun B. a car stuck in traffic C. the shaking from an eart

hquake D. X-rays from an X-ray machine at the dentist's office

1 answer:

sesenic [268]3 years ago

5 0

A Explanation: I took this test before

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How might the distance between the needle and the water droplets have affected the movement of the droplets?

shutvik [7]

Answer:

The behavior of droplets trapped in geometric structures is essential to droplet manipulation applications such as for droplet transport. Here we show that directional droplet movement can be realized by a V-shaped groove with the movement direction controlled by adjusting the surface wettability of the groove inner wall and the cross sectional angle of the groove. Experiments and analyses show that a droplet in a superhydrophobic groove translates from the immersed state to the suspended state as the cross sectional angle of the groove decreases and the suspended droplet departs from the groove bottom as the droplet volume increases. We also demonstrate that this simple grooved structure can be used to separate a water-oil mixture and generate droplets with the desired sizes. The structural effect actuated droplet movements provide a controllable droplet transport method which can be used in a wide range of droplet manipulation applications.

Explanation:

BOOM NOW I WINNNNNNNNNNNn

5 0

3 years ago

Read 2 more answers

You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C

jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0

3 years ago

A proton and an electron are released from rest, with only the electrostatic force acting. Which of the following statements mus

lbvjy [14]

Answer:

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Explanation:

The only force between the proton and the electron is electric

Fe = k q1q2 / r2

Fe = - k e2 / r2

We can see that it is an attractive force (negative sign)

The electric power energy is

U = k q1 q2 / r

U = -k e2 / r

The kinetic energy is

K = ½ m v2

With the expressions for each term we can analyze the sentences :

Their electric potential energy increases. False,

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Kinetic energy decreases False

The acceleration decreases. False, as the force increases so does the acceleration

7 0

3 years ago

When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta

nikklg [1K]

When a river flows into an ocean, it slows down and deposits materials in its delta

8 0

3 years ago

One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo

yuradex [85]

Answer:

μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

W₁ x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

cos 60 = x / L

where L is the length of the ladder

x = L cos 60

sin 60 = y / L

y = L sin60

the horizontal distance of man is

cos 60 = x2 / 7.0

x2 = 7 cos 60

we substitute

m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

fr = (735 + 2450) / 8.66

fr = 367.78 N

the friction force has the expression

fr = μ N

write the translational equilibrium equation

N - W₁ -W₂ = 0

N = m₁ g + W₂

N = 30 9.8 + 700

N = 994 N

we clear the friction force from the eucacion

μ = fr / N

μ = 367.78 / 994

μ = 0.37

3 0

3 years ago