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Dmitry [639]
3 years ago
11

Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.

3) 10 units 110 units 50 units 30 units
Physics
1 answer:
Ad libitum [116K]3 years ago
8 0

Answer:

50 units

Explanation:

Given,

  • magnitude of the first vector = a = 30 units
  • magnitude of the second vector = 70 units

As we know, From the law of vector addition,

Resultant of the addition of the two vectors is,

\therefor R\ =\ \sqrt{a^2\ +\ b^2\ +\ 2abcos\theta}

where \theta is the angle between the two vectors

And the value of cos\theat lies between -1 ≤ cos\theta ≥ 1.

For the possible values of the addition of the vectors

Therefore maximum possible value for the addition of the vector gives at the value of cos\theta\ =\ 1

\therefore R_{max}\ =\ \sqrt{30^2\ +\ 70^2\ + 2\times 30\times 70\times 1}\\\Rightarrow R_{max}\ =\ 100 units

Minimum possible value for the addition of the vectors gives at the value of cos\theta\ =\ -1

\therefore R_{min}\ =\ \sqrt{30^2\ +\ 70^2\ + 2\times 30\times 70\times (-1)}\\\Rightarrow R_{min}\ =\ 40 units

Hence the possible results of the addition of these two vectors is only 50 units which lies between the 40 units and 100 units.

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Thus, I would expect the pure water solution to freeze faster than the salt solution.
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If your friend drops a chocolate bar to you from a height of 5.0 m above your hands,
Sladkaya [172]

Answer:

<h3>1.01 s</h3>

Explanation:

Using the equation of motion S = ut+1/2gt² to solve the problem where;

u is the initial velocity of the chocolate = 0m/s

t is the time taken

g is the acceleration due to gravity = 9.81m/s²

S is the height of fall = 5.0m

Substituting the given parameter into the formula to get the time t we have;

5 = 0(t)+1/2(9.81)t²

5 = 4.905t²

t² = 5/4.905

t² = 1.019

t = √1.019

t = 1.009 secs

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A car is driven off a cliff at 39 m/s. It lands 141 m from the base. How high
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Answer: A

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4 0
2 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
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