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Anna71 [15]
3 years ago
11

Light bending as it passes through a rain drop is an example of...

Physics
2 answers:
egoroff_w [7]3 years ago
8 0
<span>C. Refraction hope this helps</span>
Dovator [93]3 years ago
6 0
The answer would be C. Refraction
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A force of 5 n produces an acceleration of 8m/s2 in mass m1 and an acceleration of 24 m/s2 in mass m2 .what acceleration would i
Inessa [10]

Acceleration of the both masses tied together= 6m/s²

Explanation:

The force is given by F= ma

so 5= m1 (8)

m1=0.625 Kg

for m2

5=m2 (24)

m2=0.208 kg

now total mass= m1+m2=0.625+0.208

Total mass=M=0.833 Kg

now F= ma

5= 0.833 (a)

a= 5/0.833

a=6m/s²

4 0
3 years ago
water is pumped from a stream at the rate of 90kg every 30s and sprayed into a farm at a velocity of 15m/s. Calculate the power
nikitadnepr [17]

Answer:

340 W

Explanation:

Power = change in energy / change in time

P = ΔKE / Δt

P = ½ mv² / Δt

P = ½ (90 kg) (15 m/s)² / (30 s)

P = 337.5 W

Rounded to 2 significant figures, the power is 340 W.

7 0
3 years ago
emperature is most closely related to which property of a liquid? (1 point) Select one: a. the volume of the liquid b. the numbe
Rainbow [258]

Answer

b. the number of atoms in each molecule.

Explanation:

5 0
3 years ago
Pleaseeeee HELPPPP THIS IS TIMED ALSO,
TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

8 0
2 years ago
Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the poten
alukav5142 [94]

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

4 0
3 years ago
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