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sergey [27]
3 years ago
13

I need help with the questions when it says what is the topic and rest

Physics
1 answer:
Wewaii [24]3 years ago
8 0
It should be a and f would I you like me to explain
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Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp
ankoles [38]

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

4 0
3 years ago
3 PHYSICAL SCIENCE: In order to figure out what is in a sealed box, you perform some tests. The box seems heavy for its size and
rjkz [21]
Some sort of magnetic metal

Metals are heavier per cubic unit than other materials such as air or water, and also are much more magnetic than other materials
7 0
4 years ago
How far will a car travel in 25 seconds at 10m/ min
Kryger [21]
Time = 25s
speed = 10m/min
= 10m / 60
= (1/6)m/s


distance = speed × time
= 25 × (1/6)
=4.167m
7 0
3 years ago
How many licks dose it take to eat a lolipop
murzikaleks [220]

Answer: Around 364 to 480

3 0
3 years ago
Read 2 more answers
Suppose a nonconducting sphere, radius r2, has a spherical cavity of radius r1 centered at the sphere's center. Assuming the cha
leva [86]

Answer:

Explanation:

a ) Between r = 0 and r = r₁

Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .

b ) From r = r₁ to r = r₂

At distance r , charge contained in the sphere of radius r

volume charge density x 4/3 π r³

q = Q x r³ / R³

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q x r³ / ε₀R³

E= Q x r / (4πε₀R³)

E ∝ r .

c )

Outside of r = r₂

charge contained in the sphere of radius r = Q

Applying Gauss's law

4πr² E = q / ε₀

4πr² E = Q  / ε₀

E = Q  / 4πε₀r²

E ∝ 1 / r² .

6 0
3 years ago
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