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2 years ago

8

A 29 cm pencil is placed 35cm in front of a convex lens and is illuminated by a spotlight. the focal point of the lens is 28cm f

rom the lens
a. what is the height of the pencils image?
b.how far from the lens will the image be focused?

1 answer:

vovikov84 [41]2 years ago

4 0

A) What is the height of the pencil image

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How far must 5N force pull a 50g toy car if 30J of energy are transferred?

Alborosie

Answer: 6 m

Explanation:

30 = 5 * d

d = 30/5

d = 6 m

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A 6000 kg lorry is reversing into a parking space at a speed of 0.5 m/s but collides with a car. The crumple zone of the car sto

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2 years ago

Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h

Bezzdna [24]

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So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..

Fn = mg - mv²/R
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2 years ago

If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

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Answer:

457.81 Hz

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From the question, it is stated that it is a question under Doppler effect.

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vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

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By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

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2 years ago

Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass

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Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

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Put the value into the formula

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$v=\dfrac{-19.375}{14.25}$

$v=-1.35\ m/s$

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2$

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Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2$

$K.E_{f}=12.98\ J$

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Using formula of loss of mechanical energy

$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0

2 years ago

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