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Lorico [155]
1 year ago
13

If an object is 16.76 cm in front of a convex mirror that has a focal length of 67.1 cm, how far behind the mirror will the imag

e appear to an observer ?
Physics
1 answer:
aniked [119]1 year ago
7 0

F = focal lenght = 67.1 cm

u= distance between mirror and object = 16.76 cm

1/f = 1/v + 1/u

1/67.1 = -1/16.76 + 1/u

1/67.1 + 1/16.76 = 1/u

u = 13.41

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A professor determined the relationship between the time spent studying (in hours) and performance on an exam.
lubasha [3.4K]
Lets write again formula for determening Ann's performance.

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3 years ago
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by
Artemon [7]

Answer:

E = 20.03 J

Explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

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The energy delivered is given by :

E=I^2Rt. ....(1)

As,

I=\dfrac{q}{t}\\\\I=\dfrac{1.67}{2}\\\\I=0.835\ A

As per Ohm's law, V = IR

R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.835}\\\\R=14.37\ \Omega

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E=0.835^2\times 14.37\times 2\\\\=20.03\ J

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6 0
3 years ago
Water flows through a garden hose which is attached to a nozzle. The water flows through hose with a speed of 1.81 m/s and throu
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Answer:

a) 17.086m

b) 0.1671 m

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Given data: speed of water through the hose  = 1.81 m/s

through the nozzle = 18.3 m/s

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a) H = v^2/2g

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= 18.3^2/(2×9.8) = 17.086 m answer

b) Again,  

H = v^2/2g

= 1.81^2/(2×9.8) = 0.1671 m

6 0
3 years ago
A projectile is fired upward at an angle θ above the horizontal with an initial speed v0. At its maximum height, what are its ve
aliya0001 [1]

Answer:

\vec{v}_{\rm max} = v\cos(\theta)(\^x)\\|\vec{v}_{\rm max}| = v\cos(\theta)\\\vec{a} = \vec{g} = -9.8\^y

Explanation:

The equations of kinematics will be used to solve this question:

y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)\\v_y = v_{y_0} + a_yt

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Therefore, the velocity vector of the projectile is

v_{max} = v_x = v\cos(\theta)

The speed of the projectile is the same.

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