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zvonat [6]
3 years ago
14

What two states of matter are pictured in the image below?

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Answer:

volume liquid

Explanation:

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Consider a constant density gas flowing steadily over an airfoil. Far upstream the velocity is V0. Halfway along the top surface
Mashutka [201]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

5 0
3 years ago
Which of the following statements is true?
hram777 [196]

All the radiation from stars is the result of nuclear fusion in their cores.

4 0
3 years ago
Read 2 more answers
"5 N, up" is an example of a ___.<br> OA) force<br> OB) mass<br> OC) weight<br> OD) magnitude
Sunny_sXe [5.5K]

Answer:

A) Force

Explanation:

It is an example of force since force is a vector quantity so it has magnitude and direction. In this case the magnitude is equal to 5 [N] and the direction is upward.

The weight can not be, as it always acts downward.

Mass is not a force, its unit is given usually in kilogram [kg]

5 0
3 years ago
An acorn with a mass of 0.300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.oo m/s?
vladimir1956 [14]

Answer: 3.75 joules

Explanation:

Given that:

Mass of acorn = 0.300 kilograms

velocity = 5.oo m/s

Kinetic energy = ?

Since, kinetic energy is the energy possessed by a moving object, its value depends on the mass M and velocity V of the acorn.

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 0.300kg x (5.00m/s)^2

= 0.5 x 0.3kg x (5.00m/s)^2

= 0.15 x (5.00m/s)^2

= 3.75 joules

Thus, the kinetic energy of the falling acorn is 3.75 joules

5 0
3 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
Delvig [45]

Answer:

E=\dfrac{\lambda }{2\pi \varepsilon _or}

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}

E_w=\dfrac{\lambda }{2\pi \varepsilon _or}

Now the electric filed due to hollow metal cylinder

E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}

E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}

Now  by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

E=\dfrac{\lambda }{2\pi \varepsilon _or}

5 0
3 years ago
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