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3 years ago

Need help ASAP

Physics

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

A, B and D

2. A

Explanation:

1.Which of the following statements about light waves is true? check all correct answers

A.They come from sources across space reason is that a light can travels form its source and its moves across the space

B.They travel in a straight line from their source reason is that the type of light that undergo these is called parallel ray of light

C.They can travel through any type of material reason not all materials light can penetrate through e.g wall

D.They can be reflected by some types of material reason is that some materals reflect light rays e.g polished mirror

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Which list is in order from smallest to biggest

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Explanation:

the solar system is smaller then nebula nebula is smaller then a galaxie and a galaxie is smaller then a universe

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A woman hikes 2.5 kilometers in 30 minutes, and then 3 kilometers in 20 minutes, and finally 2 kilometers in 10 minutes. What wa

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7.5 kph

Explanation:

2.5+3+2=7.5

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It takes a car one minute to go from rest to 30m/s east. Whqt is the acceleration of this car?

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Answer:

0.5 $m/s^{2}$

Explanation:

$a = \frac{v-u}{t} \\a = \frac{30 m/s}{60 s} \\a = 0.5 m/s^2$

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3 years ago

What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the poten

Ad libitum [116K]

Answer:

Incomplete question check attachment for complete question

Also it is given that

q1=-0.7uC

q4=-1.7uC

q3=-1.7uC

Also the distance are given as

a=2.2cm=0.022m

d2=3.6cm=0.036m

Explanation:

The potential energy due to point R is given as

The potential energy due to charge q1 and q3 plus the potential energy due to charge q4 and q1 plus the potential energy due to charge q3 and q4

So, let take it one after the other

Potential energy is give as

P.E=kq1q2/r

Therefore,

Potential energy due to charge q1 and q3

U¹³=kq1q3/r

To get the distance between charge q1 and q3, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹³=kq1q3/r

U¹³=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹³=0.254J

Potential energy due to charge q1 and q4

U¹⁴=kq1q4/r

To get the distance between charge q1 and q4, we will apply Pythagoras theorem

r=√(d2²+a²)

r=√(0.036²+0.022²)

r=0.0422m

k is a constant =9×10^9Nm²/C²

Then,

U¹⁴=kq1q4/r

U¹⁴=9×10^9×0.7×10^-6×1.7×10^-6/0.0422

U¹⁴=0.254J

Potential energy due to charge q3 and q4

U³⁴=kq3q4/r

r=2a=2×0.022=0.044m

k is a constant =9×10^9Nm²/C²

Then,

U³⁴=kq3q4/r

U³⁴=9×10^9×1.7×10^-6×1.7×10^-6/0.044

U¹⁴=0.591J

Then, the total energy is

U= U¹³+ U¹⁴ + U³⁴

U=0.254+0.254+0.591

U=1.099J

Then also, the potential energy is zero because at infinity both U¹³ and U¹⁴ will have infinite potential because their distance apart will be infinite.

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3 years ago

The field used in the Canadian football League (CFL) has the midfield marker at the 55 yard line.how long is the fiend from goal

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Answer:

110 yds

Explanation:

Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length

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