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ycow [4]
3 years ago
9

Which letter on the map represents Australia?

Physics
1 answer:
vovangra [49]3 years ago
7 0

Answer:

C because it literally in Australia but the pfp tho

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Suppose a car travels at a constant speed of 12.0 m/s. How far would it move in 3600.0 s?
beks73 [17]

Answer:

43200 m

Explanation:

speed = 12.0 m/s

time = 3600.0 s

distance = speed * time

distance = 12.0 m/s * 3600.0 s

distance = 43200 m

7 0
3 years ago
Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien
otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
  • The height at which the block stops rising is approximately 1.1415 m
  • The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

  • The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

  • Length of the incline, x_{max} = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0
2 years ago
During intercourse, Lydia constantly worries whether her partner thinks she is
Misha Larkins [42]
Answer is c spectatoring
6 0
3 years ago
Consider two wind turbines, each located in sites with the same wind speed, and each with the same efficiency. Turbine 1 has a r
Alex787 [66]

Answer:

shut up nerd

Exption:

7 0
3 years ago
A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which
Leviafan [203]

Answer:

The value is  a =  -3.7 \  m/s^2

Explanation:

From the question we are told that

   The  landing speed is  u =  71.9 \  m/s

   The  distance traveled is  d =  675 \  m

    The velocity it is reduced to is  v  =  11.3 \  m/s

   

Generally the average acceleration is mathematically represented as

      a =  \frac{ v^2  -  u^2 }{ 2 * d }

=>  a =  \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }

=>   a =  -3.7 \  m/s^2

5 0
3 years ago
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