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AlladinOne [14]
2 years ago
6

Vector A⃗ has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B⃗ has a magnitude of 3.00 and is d

irected parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C⃗ , if C⃗ =A⃗ −B⃗ .
C = 6.00; θ = 270˚
C = 3.00; θ = 270˚
C = 6.00; θ = 90˚
C = 3.00; θ = 90˚

Physics
2 answers:
Leona [35]2 years ago
5 0

The correct answer between all the choices given is the first choice, which is "<span>C = 6.00; θ = 270˚"</span><span>. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.</span>

tangare [24]2 years ago
3 0

<u>Answer</u>

C = 6.00; θ = 270˚

<u>Explanation</u>

A vector is described by giving both its magnitude and direction.

A = 3.00; 270°

B = -3.00; 90°

C = A - B. From this expression, it can be seen that we have already reversed the direction of B. So vector C is in the direction of A.

C = A - B

C = 3.00 - (-3.00)

= 6.00


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A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

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1 year ago
A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco
34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform  = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

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v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

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So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

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