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finlep [7]
4 years ago
14

When two waves undergo complete, constructive interference, which of the following increases ?

Physics
1 answer:
Lostsunrise [7]4 years ago
4 0
A speed cause it would slow down meeting eachother
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How are some types of collisions different from others?
lyudmila [28]
There are two general types of collisions, inelastic and elastic. 
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering) 
3 0
3 years ago
Students in Ms. Bailey's class were studying speed. Four students rolled balls across the floor a distance of 30 cm. Their times
Mumz [18]

Answer: C

Explanation: Ball 2 - Ball 1 - Ball 4 - Ball 3

6 0
3 years ago
Read 2 more answers
Which are examples of perfectly in elastic collisions
sweet [91]
Typical examples of inelastic collision are between cars, airlines, trains, etc.
For instance, when two trains collide, the kinetic energy of each train is transformed into heat, which explains why, most of the times, there is a fire after a collision. However, the momentum of the two trains that are involved in the collision remains unaffected. So, the trains collide with all their speed, maintaining their momentum, yet their kinetic energy is transformed into heat energy.
Another way to explain a train or a car collision is this: when the two trains or cars collide, they stick together while slowing down. They slow down because their kinetic energy is gradually lost. Still, they collide because they conserve their momentum.
8 0
3 years ago
A 975-kg pickup comes to rest from a speed of 87.5 km/h in a distance of 125 m. Suppose the pickup is initially traveling in the
Savatey [412]

Answer:

A) Force = 2303.925 N in the negative x-direction

B) F ≈ 143998.28 N

C) Ratio = 62.5

Explanation:

A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.

Let's find the deceleration using Newton's law of motion formula;

v² = u² + 2as

where;

v = final velocity,

u = initial velocity,

s = displacement

a = acceleration

We are given;

u = 87.5 km/h = 24.3056 m/s

s = 125 m

v = 0 m/s

Thus;

0 = (24.3056)² + 2a(125)

- (24.3056)²= 250a

a = - 24.3056²/250

a = - 2.363 m/s²

Now, force = mass × acceleration

We are given mass = 975 kg

Thus;

Force = 975 x (-2.363)

Force = 2303.925 N in the negative x-direction

B) formula for kinetic energy is

KE = ½mv²

KE = ½(975)(24.3056)²

= 287996.568288 J

Work done on impact = F x 2

Thus;

2F = 287996.568288

F = 287996.568288/2

F ≈ 143998.28 N

C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5

3 0
3 years ago
the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab
JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

  • <em>Mass of the first block, m₁ = 4.0 kg</em>
  • <em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

m_2 g - T = m_1 a

where;

  • <em>T </em><em>is the tension in the connecting string due weight of the first block</em>

m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

5 0
3 years ago
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