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Valentin [98]
3 years ago
13

if an object 18mm high is placed 12mm from a diverging lens and the image is formed 4mm in front of the lens what is the height

of the image
Physics
1 answer:
brilliants [131]3 years ago
5 0

To solve this problem we use an amplification formula for divergent lenses

M = \frac{i}{o} = \frac{h'}{h}

Where:

i: distance of the image to the lens

o: Distance from the object to the lens

h = height of the object

h '= height of the image

M = \frac{4mm}{12mm} = \frac{h'}{18mm}

h'= 18mm\frac{4mm}{12mm}

h '= 6 mm

The height is 6 mm

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Single Displacement Reaction Definition. A single displacement reaction is a chemical reaction where one reactant is exchanged for one ion of a second reactant. It is also known as a single replacement reaction.

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3 years ago
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Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?
IRISSAK [1]

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

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3 years ago
What is the interaction between surface water and groundwater in a watershed?
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Answer:

Groundwater occasionally discharges into surface water and then, they flow together as a body of water in a watershed.

Explanation:

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Therefore, when groundwater discharges into a body of surface water, for example a stream, the stream just like several other streams in a watershed would flow into a larger body of water.

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3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

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If a cat can exert 2000N Of Force to move a trailer 50m is 20 seconds how much power did the car use ?
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