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2 years ago

An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance

Physics

1 answer:

Tomtit [17]2 years ago

3 0

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

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

By rearrangement and making (v) the subject of the above formula:

$v = \dfrac{uf}{u-f}$

replacing the given values:

$v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}$

$v = \dfrac{-3.0}{(-3.5)}$

v = 0.857 cm

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Passengers on an airplane move from rest to 86 meters per second before the airplane takes off. If the airplane takes 100 second

VikaD [51]

Utilize the formula: $V _{f} = V _{i} + a\Delta t$

$V _{f}$ = Final Velocity (86 m/s)

$V _{f}$ = Initial Velocity (0 m/s)

a = acceleration (m/s²)

$\Delta$ t = Time (100 seconds)

As a result,

86 m/s = 0 + (a)(100 seconds)

Using algebra, divide 86 m/s by 100 seconds:

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3 years ago

Part 1 :

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

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3 years ago

When you eat a large meal and your body absorbs a lot of glucose and that makes its way to the interstitial fluid before going i

Agata [3.3K]

Answer:A. It is modified by the cell, so there is still more glucose on the outside of the cell than inside it

Explanation: The cells are the primary unit of a living organism,it is the place where cellular respiration takes place, which is the use of oxygen to break down food substances such as Glucose to release energy in the form of Adenosine trisphosphate (ATP) and give out Carbondioxide as a by-product of cellular respiration.

Most of the Glucose enter the cells to get modified through cellular respiration so their is still more Glucose outside than the Glucose inside.

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2 years ago

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive

Irina-Kira [14]

Answer:

Explanation:

In the first case you can use the expression for the Doppler effect when the source is getting closer and getting away

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f' = perceived frequency when the source is getting closer

f'' = perceived frequency when the source is getting away

f = source frequency

v = relative speed

vs = sound speed

by dividing (1) and (2) you have

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but this is the relative velocity, you have that

$v=v_{sir}-v_{car}\\v_{sir}=v+v_{car}=67.5\frac{m}{s}+35\frac{m}{s}=102.5\frac{m}{s}$

a. hence, the speed of the police car is 102.5m/s

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3 years ago

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Answer:

0.45 m/s

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