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1 year ago

100 Points!!!

1 answer:

4 0

(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.

(2) When light rays reflect, they bounce back.

(3) Images formed by a concave lens will look magnified.

(4) When light rays enter a different medium, they bend.

<h3>1.0 Object placed further from the lens than the focal point</h3>

The image of an object placed further from the lens than the focal point will be diminished and inverted.

Thus, the correct answer will be "upside down and smaller than the object".

<h3>2.0 What is reflection of light?</h3>

The ability of light to bounce back when it strike a hard surface is known as refection.

<h3>3.0 Image formed by concave lens</h3>

A concave lens is diverging lens is usually virtual, erect and magnified.

<h3>4.0 Refraction of light</h3>

The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.

Learn more about reflection and refraction of light here: brainly.com/question/1191238

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Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

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$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

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<u>Answer:</u>

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