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grandymaker [24]
3 years ago
11

How deep is the outer core beneath the surface

Physics
1 answer:
sukhopar [10]3 years ago
6 0
Precisely around 1,800 miles below.
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How are the concepts of impulse and momentum related?
butalik [34]

Answer:

The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. ... The collision would change the halfback's speed and thus his momentum.

Explanation:

4 0
3 years ago
A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p
Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

v_x = v cos45

in y direction

v_y = vsin45

now displacement in x direction

x = (vcos45)t + 0

displacement in y direction

y = (vsin45)t - \frac{1}{2}gt^2

now from above two equations we have

y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2

y = xtan45 - \frac{1}{2v^2cos^245}gx^2

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

<em>C. parabolic curve.</em>

8 0
3 years ago
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
4 years ago
Match the situation with the energy transformation ITEMBANK: Move to Top A boy shooting a rubber band across the classroom A chi
Sonbull [250]
A boy shooting a rubber band across the classroom --> 
Elastic potential energy transformed into kinetic energy 
<span>The initial energy is the energy stored in the muscles of the boy's arm, which is elastic potential energy. This is converted into motion of the rubber, therefore kinetic energy

A child going down a slide on a playground --> </span>Gravitational potential energy transformed into kinetic energy 
On top of the slide, all the energy of the child is gravitational potential energy due to its height with respect to the ground (E=mgh). when it moves down the slide, this is converted into kinetic energy, because the child acquires a speed v (E=1/2 mv^2)
<span>
Rubbing your hands together to warm them on a cold day --> </span>Kinetic energy being transformed into thermal energy <span>
When rubbing hands, we are moving them (kinetic energy), and this energy raises the temperature of the hand's surface (thermal energy)

Turning on a battery operated light --> </span>
Chemical potential energy transformed into radiant energy <span>
A battery works by mean of chemical reactions (chemical potential energy), producing light (so, emitting energy by radiation, i.e. radiant energy)

Using a dc electric motor --> </span> Electrical energy transformed into kinetic energy<span>
A dc electric motor works using  currents (so, electrical energy), and the energy produced can be used for example to accelerate a car (kinetic energy)

Using a gas power heater to warm a room --> </span>Chemical potential energy transformed into thermal energy 
<span>A gas power heater burns gases (so, chemical reaction, i.e. chemical potential energy) to raise the temperature of the room (thermal energy)

Using a hand crank generator to produce electric current --> Kinetic energy transformed into electrical energy
In a hand-crank generator, the handle is being rotated (kinetic energy) in order to produce an electric current (electrical energy)

Using the light in your room that is plugged into the wall --> </span>Electrical energy transformed into radiant energy  
<span>The lamp works by using electrical current flowing into a resistor (electrical energy) and it produces light, so it emits energy by electromagnetic radiation (radiant energy)




</span> <span>

</span>
3 0
3 years ago
What will occur when the waves peak at the same place at the same time?
dedylja [7]
Hello!

This is a matter of superposition.
When the waves peak at the same time and place, they produce constructive interference, meaning the waves interact together in a positive way, to make a wave with Amplitude of both waves added together. When the peaks differ however, at the same time and place, then it is destructive interference and the waves essentially cancel each other out.

Hope this helps. Any questions please just ask. Thank you kindly.
5 0
3 years ago
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