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Fiesta28 [93]
2 years ago
9

Without his glasses, Isaac can see objects clearly only if they are less than 4.7 m from his eyes.

Physics
1 answer:
Mazyrski [523]2 years ago
6 0

Answer:

The focal length is  f  =  -4.721 \  m

Explanation:

From the question we are told that

  The  image distance  is  v  =  4.7 +  (2.1 *10^{-2}) =  - 4.721 \ m the negative means  image is formed behind the lens

Generally the object distance is  u  =  \infty this mean the light ray coming from the object are parallel to each other

So from lens equation

       \frac{1}{\infty } -   \frac{1}{4.721} = \frac{1}{f}

=>   f  =  -4.721 \  m

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Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0. 250 kilograms, what should be
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Sum of all forces = mass * acceleration

Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)

Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
Ft= 2.45N

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2 years ago
If you jog at a speed of 1.5m/s for 20 seconds how far di you travel
Harlamova29_29 [7]

Answer: 30m

Explanation:

Given:

Speed: 1.5m/s

Time: 20 seconds

Distance = speed × time

Distance = 1.5 × 20

= 30m

Therefore you will travel 30m

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2 years ago
A sports car traveling at 24.7 m/s slows at a constant rate to a stop in 16.00 s. What is the displacement of the sports car in
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Explanation:

Displacement=Velocity×time

=24.7×16.00

=395.2m

Therefore the displacement within the time interval is 395.2m

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A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou
zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius  = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}

E=6.88\times10^{7}\ N/C

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d).  For, r = 7.00 cm

The electric field is

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}

E=5.43\times10^{6}\ N/C

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0
3 years ago
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