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Dmitry_Shevchenko [17]
2 years ago
12

Please Help Me!

Physics
1 answer:
balandron [24]2 years ago
5 0

The image distance when a boy holds a toy soldier in front of a concave mirror, with a focal length of 0.45 m. is -0.56 m.

<h3>What is image distance?</h3>

This is the distance between the image formed and the focus when an object is placed in front of a plane mirror.

To calculate the image distance, we use the formula below.

Formula:

  • 1/f = 1/u+1/v........... Equation 1

Where:

  • f = Focal length of the mirror
  • v = Image distance
  • u = object distance

From the question,

Given:

  • f = 0.45 m
  • u = 0.25 m

Substitute these values into equation 1 and solve for the image distance

  • 1/0.45 = 1/0.25 + 1/v
  • 2.22 = 4+1/v
  • 1/v = 2.22-4
  • 1/v = -1.78
  • v = 1/(-1.78)
  • v = -0.56 m

Hence, The image distance is -0.56 m.

Learn more about image distance here: brainly.com/question/17273444

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Explanation:

A simple harmonic oscillation is an oscillator that is neither driven nor damped.

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In grinding a steel knife blade (specific heat = 0.11 cal/g-c),the metal can get as hot as 400C. If the blade has a mass of 80g,
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Answer:

33 g.

Explanation:

Assuming no heat transfer can be possible except for heat exchange between water and steel, we can say that the heat lost by the knife, must be equal to the heat gained by the water.

As we have a limit for the maximum temperature of both elements (once reached a final thermal equilibrium), of 100ºC, which means that the maximum allowable change in temperature will be of 300º C for the knife, and of 80º C for the water.

Empirically , it has been showed that for a heat exchange process using only conduction, the heat needed to raise the temperature of a body, is proportional to the mass, being the proportionality constant a factor that depends on the material, called specific heat.

So, we can write the following equation:

cs*mk*Δtk = cw*mw*Δtw

Replacing by the givens of the question, we have:

0.11 cal/gºC * 80 g * 300ºC = 1 cal/gºC*mw*80ºC

Solving for mw = 2,640 cal / 80 cal/g =33 g.

5 0
3 years ago
A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.
34kurt

To solve this problem it is necessary to apply the equation related to the Gravitational Force, the equation describes that

F = \frac{GMm}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth (or Bigger star)

m = Mass of Object  (or smallest star)

r = Radius

From the statement we know that once the impact is made, the golf ball is subjected to the forces that are exerted in nature. Since the air resistance, which would represent the drag force, is ignored. Only the forces related to gravity remain.

The gravitational force carries 'pushes' or 'attracts' the body towards the earth, while the speed decreases as it reaches its maximum height.

When the ball has reached its maximum height only the force of gravity begins to act on it, generating the attraction to the earth in parabolic motion.

Therefore the correct answer is B.

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