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Pani-rosa [81]
2 years ago
10

The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed o

f an object located 63 cm to the left of the lens on the optical axis?
.... cm to the left of the lens

....What is the magnification of the image?
Physics
1 answer:
Arada [10]2 years ago
7 0

The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

<h3>Location of the image</h3>

The image distance formed by the lens is calculated as follows;

1/v + 1/u = -1/f

where;

  • v is the image distance
  • u is the object distance
  • f is the focal length of the lens

1/v = -1/f - 1/u

1/v = -(-1/26) - 1/63

1/v = 1/26 - 1/63

1/v = 0.022588

v = 1/0.022588

v = 44.27 cm

<h3>What is magnification of lens?</h3>

The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

<h3>Magnification of the image formed</h3>

The magnification of the image is calculated as follows;

Magnification = image distance/object distance

M = 44.27/63

M = 0.7

Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

Learn more about magnification here: brainly.com/question/1599771

#SPJ1

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A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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When a car slows down suddenly, passengers in the car tend to move toward the front of the car. What is this due to?
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Hey there Kendrell!

Yes, this is very true, when the car slows down, our bodies will tend to lean forward a little bit, and this is actually due to the "motion of inertia".

Inertia allows for this to happen, this is why in this case, we have this case.

Hope this helps.
~Jurgen


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The statement “Heavy objects fall faster than light objects” is an example of a(n) _______.
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D. Theory is the answer
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A point charge of 5.0 x 10^-7 C moves to the right at 2.6 x 10^5 m/s in a magnetic field that is directed into the screen and ha
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The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

<h3>What is magnetic force?</h3>

A magnetic force is the force that act in a magnetic field.

To calculate the magnetic force, we use the formula below.

Formula:

  • F = qvB.........Equation 1

Where:

  • F = magnetic force
  • q = point charge
  • v = Velocity of the the charge
  • B = Field strength

From the question,

Given:

  • q = 5.0×10⁻⁷ C
  • v = 2.6×10⁵ m/s
  • B = 1.8×10⁻² T

Substitute these values into equation 2

  • F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
  • F = 23.4×10⁻⁴
  • F = 2.34×10⁻³ N

Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

Learn more about magnetic force here: brainly.com/question/2279150

#SPJ12

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Describe how two isotopes of nitrogen differ from two nitrogen ions???PLEASE ANSWER 20 POINTS
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