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Leto [7]
3 years ago
8

A ray diagram is shown.

Physics
1 answer:
alexandr402 [8]3 years ago
7 0

Answer:

  Z

Explanation:

The purple and yellow rays from the same point on the object are shown converging at point Z, where the image will be.

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A small plane starts from rest and accelerates uniformly to the east to a takeoff velocity of 70 m/s in 5 seconds. What is the p
Rainbow [258]
Hope this helps, if you need clarification i got you

3 0
2 years ago
An ideal gas is at a pressure 1.00 Ã 105 n/m2 and occupies a volume 2.00 m3. if the gas is compressed to a volume 1.00 m3 while
blsea [12.9K]
The behavior of an ideal gas at constant temperature obeys Boyle's Law of
p*V = constant
where
p = pressure
V = volume.

Given:
State 1:  
  p₁ = 10⁵ N/m² (Pa)
  V₁ = 2 m³
State 2:
  V₂ = 1 m³

Therefore the pressure at state 2 is given by
p₂V₂ = p₁V₁
or
p₂ = (V₁/V₂) p₁
    = 2 x 10⁵ Pa

Answer: 2 x 10⁵ N/m² or 2 atm.
4 0
3 years ago
A boulder released by a rockslide rolls 60 m in 4 s. What is its average speed?
Gelneren [198K]

Answer:

15 m per second

900m per minute

54,000 per hour

Explanation:

60 divided by 4 to get per second then times 60 for per minutes

then times 60 to get per hour

5 0
3 years ago
A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releas
Sever21 [200]
Thank you for posting your question here at brainly. Below is the solution:

Ke up top = 1/2*.25 *225 
<span>gain of Pot energy = .25*9.81*1.2 </span>
<span>work input = (1/4)(2 pi *.6)*30 </span>

<span>so </span>
<span>sum of those 3 energies = </span>
<span>(1/2)(.25)v^2</span>
3 0
3 years ago
Simple physics question, check the document. Should take about 3-5 minutes.
Ahat [919]

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

5 0
3 years ago
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