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3 years ago

A lens uses what process to deflect light rays passing through it? a. reflection b. refraction c. absorption d. transparency

Physics

2 answers:

USPshnik [31]3 years ago

8 0

A lens uses the process of refraction to deflect light rays from passing through it.

Doss [256]3 years ago

3 0

Answer:

I know this is from 2 years ago.

Explanation:

But I'm trying to get a challenge done, and it's B. refraction.

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In beachville, the last high tide occurred at 1:00 p.m. the next high tide will happen around _____. 1:25

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At Around 1:25 a.m. it will happen

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3 years ago

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Can someone pls answer number 8 for me pls

fomenos

Answer:

1.33 m

Explanation:

Tom weighs 600 N and is 2 meters from the pivot. So his moment is:

τ = Fr

τ = (600 N) (2 m)

τ = 1200 Nm

To balance the see-saw, Jeremy must create an equal and opposite moment.

τ = Fr

1200 Nm = (900 N) r

r = 1.33 m

So Jeremy must sit 1.33 m from the pivot on the other side.

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3 years ago

A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak

shusha [124]

Answer:

17.3 m

Explanation:

Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

$y=y_0+v_ot +\dfrac{1}{2}gt^2$

Putting all the values,

$0=18.2+6.69t-\dfrac{1}{2}\times 9.8t^2\\\\-4.9t^2+6.69t+18.2=0\\\\t=\dfrac{-6.69+\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}, \dfrac{-6.69-\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}\\\\t=-1.36\ s\ \text{and}\ t=2.72\ s$

Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

$d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m$

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3 years ago

How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and

aev [14]

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h ( g = 10 m/s²)

h= 46.66 m

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3 years ago

Carbon and hydrogen are examples of pure substances or

Ede4ka [16]

The are elements on the periodic table

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