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Advocard [28]
1 year ago
15

consider the points ​p(​,​,​) and ​q(​,​,​). a. find and state your answer in two​ forms: and aibjck. b. find the magnitude of .

c. find two unit vectors parallel to .
Physics
1 answer:
astra-53 [7]1 year ago
6 0

Consider the points <u>p(,,) and q(,,):</u>

P (-4,1,0)

Q (-4,-5,2)

A. find and state your answer in two forms

PQ = <u>(-4+4, -5-1, 2-0)</u>

= <u>(0,-6,2)</u>

B. find the magnitude

 |PQ|= √(0)2+ (-6)2 + (2)2

= √36+4

= √40

= <u>2√10</u>

C. find two unit vectors

Unit vector parallel to PQ

= < 0, -3/√10, 1/√10 >

= <u>< 0, 3/√10, -1/√10 ></u>

In physics, magnitude is defined as the maximum size and orientation of an object. Size is used as a common factor for vector size and scalar size. By definition, we know that a scalar quantity is one that has only one quantity.

Force Magnitude is a numerical value that expresses the strength of a force. Example: Suppose the force is = 10 N towards the east. "East" is the direction, "10" is the strength.

Learn more about magnitude here:- brainly.com/question/24468862

#SPJ4

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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!!!

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