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PIT_PIT [208]
3 years ago
13

A ray diagram shows that an object is placed between the focal point and the vertex of a convex mirror. What are the characteris

tics of the image produced by the object? inverted, smaller than object, real upright, smaller than object, virtual inverted, larger than object, real upright, larger than object, virtual

Physics
2 answers:
Aleonysh [2.5K]3 years ago
7 0
The correct answer is: <span>smaller than object, virtual and upright.
</span>
When the object is placed between the focal point and the vertex of a convex mirror, then the image formed will be <span>located behind the convex mirror. It will be virtual image and an upright image but reduced in size (i.e., smaller than the object)</span>

olga2289 [7]3 years ago
4 0

On Edge, the answer is<u> B.</u>

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What is one property of iodine
mamaluj [8]

Answer: I2

Explanation:

6 0
3 years ago
Assuming this rate to be constant how many years will pass before the radius of the Moon's orbit increases by 3.6 x 10^6 m
Evgen [1.6K]

The number of years that will pass before the radius of the Moon's orbit increases by 3.6 x 10^6 m will be 90000000 years.

<h3>How to compute the value?</h3>

From the information given, the orbit of the moon is increasing in radius at approximately 4.0cm/yr.

Therefore, we will convey the centimeters to meter. This will be 4cm will be:

= 4/100 = 0.04m/yr.

Time = Distance / Speed

Time = 3.6 x 10^6/0.04

Time = 90000000 years.

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brainly.com/question/13262798

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Complete question:

Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the moon is increasing in radius at approximately 4.0cm/y. Assuming this rate to be constant how many years will pass before the radius of the Moon's orbit increases by 3.6 x 10^6

6 0
2 years ago
What did Thomson's and Rutherford's experiments have in common?
sdas [7]
<span>They both used charged particles in their experiments.</span>
4 0
3 years ago
Read 2 more answers
A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the
HACTEHA [7]

The work done onto the car is 506,250 J

The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.

From the parameters given:

  • The mass of the car = 1500 kg
  • The initial speed = 30 m/s
  • The final speed = 15 m/s

The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)

\mathbf{=\dfrac{1}{2} mv_1^2 -\dfrac{1}{2} mv_2^2}

\mathbf{=\dfrac{1}{2} m(v_1^2 - v_2^2)}

\mathbf{=\dfrac{1}{2} \times 1500 \times (30^2 - 15^2)}

= 506,250 J

Therefore, we can conclude that the work done on the car is 506,250 J

Learn more about work done here:

brainly.com/question/18762601

7 0
3 years ago
A ship leaves a port at noon and travels due west at 20 knots. At 6 PM, a second ship leaves the same port and travels northwest
Amanda [17]

Answer:

v = 12.44 Knots

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

so after time "t" of 2nd ship motion the two ships positions are given as

r_1 = 20(t + 6)\hat i

r_2 = 15(t)(cos45\hat i + sin45\hat j)

now we can find the distance between two ships as

x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}

now we have

x^2 = (120 + 9.4 t)^2 + (10.6 t)^2

x^2 = 200.72 t^2 + 14400 + 2256 t

now we will differentiate it with respect to time

2x\frac{dx}{dt} = 401.44 t + 2256

here we know that

t = \frac{90}{15} = 6 hours

so we have

x = 187.5

now we have

2(187.5) v = 401.44(6) + 2256

v = 12.44 Knots

5 0
3 years ago
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