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Ainat [17]
2 years ago
10

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

ification produced by the mirror.
Physics
1 answer:
77julia77 [94]2 years ago
7 0

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

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A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car
Furkat [3]

Answer:

Part a)

a = 4.86 m/s^2

Part b)

d = 155.52 m

Part c)

v_f = 48.6 m/s

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

v_f = 87 mph = 38.88 m/s

now we have

Part a)

acceleration is rate of change in velocity

a = \frac{v_f - v_i}{t}

a = \frac{38.88 - 0}{8}

a = 4.86 m/s^2

Part b)

distance moved by car with uniform acceleration is given as

d = \frac{v_f + v_i}{2} t

d = \frac{38.88 + 0}{2} 8

d = 155.52 m

Part c)

As we know that the car start from rest

so final speed after t = 10 s

v_f = v_i + at

v_f = 0 + (4.86)10

v_f = 48.6 m/s

3 0
3 years ago
Which of the following statements is true?
inn [45]
C is the right answe
5 0
2 years ago
The Balmer series is formed by electron transitions in hydrogen that
yulyashka [42]

Answer:

b. end on the n = 2 shell.

Explanation:

When hydrogen atoms move from higher energy level to lower energy level then it shows spectral lines and these lines are known as Balmer series. The only four lines are visible and other liens are not in the visible range.

The Balmer series formed by hydrogen electron and it ends when n = 2.

Therefore the answer is b.

b. end on the n = 2 shell.

8 0
3 years ago
A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times
Ira Lisetskai [31]

Answer:

9R

Explanation:

We know that the resistance is R=\rho *\frac{L}{A}.

If we stretch the wire to a new length L2 = 3L, the cross-sectional area will also change. If the cross-sectional area doesn't change throughout the wire, we can say that:

Volume = L*A = 3L * A2    being A2 the new area after stretching the wire.

Since the volume remains the same we conclude that A2 = A/3

With this information, we calculate the new resistance:

R2=\rho *\frac{L2}{A2}=\rho *\frac{3*L}{A/3}=\rho * 9 * \frac{L}{A}

Since R=\rho *\frac{L}{A}, and by simple inspection of the previous equation, we get:

<em>R2 = 9*R</em>

6 0
3 years ago
Spurs"" are probably the result of ____.
egoroff_w [7]

Spurs are probably the result of <u>self-sustaining</u> <u>star formation.</u>

<h3>What is the formation of gaseous spurs in spiral galaxies?</h3>

The gigantic form of the magnificent doppelganger spiral patterns that spiral outward from the galactic cores gave spiral galaxies their name. These light arms of spiral galaxies are frequently seen in optical pictures to be speckled with bright star-forming areas at regular intervals.

Smaller structures spread forth and rearward into the interarm area from each major spiral arm. Spiral-arm also known as spurs are the name given to these substructures. Sometimes the spurs are also filled with star-forming clusters. As a consequence, we may draw the conclusion that spurs most likely emerge from self-sustaining star formation.

Learn more about the spiral galaxies here:

brainly.com/question/13956361

#SPJ11

5 0
1 year ago
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