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3 years ago

A 6.00 cm tall light bulb is placed a distance of 54.2 cm from a double convex lens

Physics

1 answer:

BigorU [14]3 years ago

7 0

Answer:

1 / f = 1 / i + 1 / o thin lens equation

1 / i = 1 / f - 1 / o = (o - f) / (o * f)

i = o * f / (o - f)

i = 54.2 * 12.7 / (54.2 - 12.7) = 16.6 cm image distance

Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall

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What’s Chemical energy

AVprozaik [17]

Answer:

Chemical energy is energy that chemical substance to undergoes a chemical reaction to make a new substance.

Explanation:

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3 years ago

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peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie pla

zavuch27 [327]

Explanation:

It is given that,

Diameter of the peach pie, d = 9 inches

Radius of the pie, r = 4.5 inches

The tray is rotated such that the rim of the pie plate moves through a distance of 183 inches, d = 183 inches

Let $\theta$ is the angular distance that the pie plate has moved through.

It is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{183}{4.5}$

$\theta=40.66\ radian$

Since, 1 radian = 57.29 degrees

$\theta=2329.64\ degrees$

Since, 1 radian = 0.159155 revolution

$\theta=6.47124\ revolution$

Hence, this is the required solution.

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3 years ago

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou

mojhsa [17]

Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

$g = \dfrac{GM}{r^2}$

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3 years ago

An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car

saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

$F = ma$

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

$-10000 =2000\times a$

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

$a = -\dfrac{10000}{2000}$

$a=-5\ m/s^2$

For the distance,

Using third equation of motion

$v^2-u^2=2as$

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

$0-30^2=2\times(-5)\times s$

$s = 90\ m$

Hence, The car traveled the distance before stopping is 90 m.

6 0

3 years ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

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4 years ago