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slega [8]
3 years ago
6

A local fun house incorporates a gently curved, concave, spherical mirror into its display. When a child stands 1.2 m from the m

irror, her reflection is upside down and appears to float in front of the mirror. When she stands about 0.8 m from the mirror, she sees only a blur reflected in the mirror, but when she stands about 0.5 m from the mirror, her reflection is right side up and appears to be behind the mirror. The approximate focal length of the mirror is
Physics
1 answer:
lys-0071 [83]3 years ago
3 0

Options:

a) more than 0.8 m .

b) equal to 0.8 m .

c) between 0.5 m and 0.8 m .

d) less than 0.5 m .

Answer:

b) equal to 0.8 m .

Explanation:

Note:

An upside down image = Inverted Image

An image that appears in front of the mirror = Real image

An image that appears behind the mirror = Virtual image

Let the object distance from the pole of the mirror be u

When the child stands 1.2 m from the mirror:

u = 1.2 m ( Real and Inverted image of the child is formed)

When the child stands about 0.8 m from the mirror:

u = 0.8 m (Virtual, erect and magnified image of the child is formed)

When the child stands about 0.5 m from the mirror:

u = 0.5 m ( Virtual and erect image of the child is formed)

Note: All objects positioned behind the focal length of a concave mirror are always real. Objects start becoming virtual when they are placed on the focal length or in front of it (Close to the pole of the mirror), although objects placed on the focus has its image formed at infinity.

Since the nature of the image formed changed from real to virtual when the child stands about 0.8 m from the mirror, then the focal length is approximately equal to 0.8 m

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Answer:According to the Equation (2), centripetal force is proportional to the square of the speed for an object of given mass M rotating in a given radius R.

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
2 years ago
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