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Masja [62]
3 years ago
9

Which of these is described as force per unit charge on a positive test charge?

Physics
1 answer:
Nuetrik [128]3 years ago
4 0
I'm 99% sure that it's "D. Electric Field"
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What are some ways to defend a castle against attack?
ella [17]

Answer:

you can build a mote around your castle and put crocodiles in it.

Explanation:

they will have to go in the water to get to your castle and will get eaten.

4 0
2 years ago
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A pencil sharpener has a handle with a radius of 1.5 cm. The sharpening mechanism has a radius of 0.5 cm. What is the IMA of the
abruzzese [7]
1.5 / 0.5 = 3 I believe this is the right answer
8 0
3 years ago
A certain cloud contains 220 water droplets per cubic centimeter. If 1cubic meter = 1,000,000cubic centimeters, how many drops a
lianna [129]
Wouldn't you just have to multiply 220 by 1,000,000? That would mean there are 220,000,000 water droplets in one cubic of the cloud.
3 0
3 years ago
g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest
ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

3 0
2 years ago
Estimate the length of the neptunian year using the fact that the earth is 1.50Ã108km from the sun on average.
Ksenya-84 [330]

Answer:

Explanation:

We know that,

Neptune is 4.5×10^9 km from the sun

And given that,

Earth is 1.5×10^8km from sun

Then,

Let P be the orbital period and

Let a be the semi-major axis

Using Keplers third law

Then, the relation between the orbital period and the semi major axis is

P² ∝ a³

Then,

P² = ka³

P²/a³ = k

So,

P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³

Period of earth P(earth) =1year

Semi major axis of earth is

a(earth) = 1.5×10^8km

The semi major axis of Neptune is

a (Neptune) = 4.5×10^9km

So,

P(E)²/a(E)³ = P(N)² / a(N)³

1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³

Cross multiply

P(N)² = (4.5×10^9)³ / (1.5×10^8)³

P(N)² = 27000

P(N) =√27000

P(N) = 164.32years

The period of Neptune is 164.32years

6 0
3 years ago
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