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3 years ago

A book 15.6 cm tall is 28.7 cm to the left of a plane mirror. What is the height of the image formed?

Physics

1 answer:

Ira Lisetskai [31]3 years ago

3 0

Answer:

15.6cm

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Which is an example of conduction?

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A hand is burned when touching a pot handle

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3 years ago

Read 2 more answers

20 POINTS 20 POINTS On a bright sunny day you decide to take a walk. You begin at your home and walk 1000 meters to an ice cream

Kay [80]

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

4 0

3 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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3 years ago

Which fossil fuel is produced as a by-product that occurs when bacteria decompose organic material under anaerobic conditions?

Mumz [18]

Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.

<h3><u>Explanation:</u></h3>

Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.

The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.

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3 years ago

A book is to be produced with pages of thickness 0.125mm.If the book is to be exactly 1cm thick what is the maximum number of pa

miskamm [114]

Answer:

1cm=10mm

10cm/0.125mm

80pages

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3 years ago