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andrezito [222]
2 years ago
12

Write a sentence on how these words are used in real life situations

Physics
1 answer:
ipn [44]2 years ago
6 0

Answer:

Explanation:

You can approach an expression for the instantaneous velocity at any point on the path by taking the limit as the time interval gets smaller and smaller. Such a limiting process is called a derivative and the instantaneous velocity can be defined as.#3

For the special case of straight line motion in the x direction, the average velocity takes the form: If the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as For this special case, these expressions give the same result. Example for non-constant acceleration#1

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Which shows a decrease in fluid pressure? A. A fan is turned from high speed to low speed. B. Oxygen is compressed as it is put
Volgvan

Answer:

Option A.

A fan is turned from high speed to low speed.

Explanation:

It is important to note that air is also a fluid.

In a system, static pressure of air increases with the speed of rotation of the fan. This is because when the speed of the fan is increased, the force with which it is pushing the air molecules is increased. Since pressure is a relationship between force and area, the pressure of the air molecules will be increased.

Conversely, when the speed of the fan is reduced, the priming force on the air molecules will be reduced, hence the pressure of the air will drop.

This makes option A the correct option

8 0
3 years ago
Read 2 more answers
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
2 years ago
What happens when the falling object reaches its terminal velocity? It will stop _________
lora16 [44]

Answer:

LOL i don't even know

Explanation:

7 0
2 years ago
Buoyancy increases with the increase in the density of a) Submerged body b) Fluid​
AlekseyPX

Answer:

Submerged body

Explanation:

  • If buoyancy is greater than weight then object will float.
  • If buoyancy is less then weight then object will sink.
  • If buoyancy=weight then objects remains stable
4 0
3 years ago
The number of planets, besides the earth, that are visible to the unaided eye is
sergejj [24]
The answer should be 5 it’s easy
3 0
3 years ago
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