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sleet_krkn [62]
2 years ago
8

A block of height 120mm is placed on top of another block with a height of 1.50 m. What is the height of the two blocks together

? (1000mm = 1m)
a) 1.72 m

b) 1.62 m

c) 2.70 m

d) 121.5 m
Physics
1 answer:
Kamila [148]2 years ago
4 0
We will first convert all units to meters and then solve the problem.
We are given that:
1000 mm = 1 m
120 mm = ?? meters
using cross multiplication:
120 mm = (120*1) / 1000 = 0.12 m

Now, when the two objects are placed over each other, their total height is the result of summation of both heights, therefore:
total height = 0.12 + 1.5 = 1.62 m

Based on the above calculations, the correct choice is:
b) 1.62 m 
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Choice C.
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TiliK225 [7]

The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 8.5 m/s
  • Final velocity (v) = 7.5 m/s
  • Time (t) = 5 ms = 0.25 s
  • Mass (m) = 2.5 Kg
  • Force (F) = ?

<h3>How to determine the force</h3>

The force exerted on the ball can be obtained as follow:

F = m(v + u) / t

F = [2.5(7.5 + 8.5)]/ 0.25

F = 40 / 0.25

F = 160 N

Thus, the force exerted on the ball is 160 N

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2 years ago
Describe a process in which energy changes forms at least twice
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9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku
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Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

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Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

  • v₁ = -(\frac{\sqrt{2} }{2} × 3.5)·i + (\frac{\sqrt{2} }{2} × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

  • v₂ =  (\frac{\sqrt{2} }{2} × 4.0)·i - (\frac{\sqrt{2} }{2} × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-(\frac{\sqrt{2} }{2} × 3.5) - (\frac{\sqrt{2} }{2} × 4.0))·i + ((\frac{\sqrt{2} }{2} × 3.5 + 12.5) + (\frac{\sqrt{2} }{2} × 4.0 + 15))·j

  • v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

Red Skull's velocity relative to Captain America,  v₁₂ = -\frac{ 15 \cdot \sqrt{2} }{4}·i + \frac{ 110 + 15 \cdot \sqrt{2} }{4}·j

  • v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

  • Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

  • The direction is arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}

Therefore;

  • Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

  • |v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

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If a moving object travels for a distance of 160 m. In 20 seconds what’s the average speed
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Answer:

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