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Aleks [24]
4 years ago
7

Explain the term in physics viscosity​

Physics
2 answers:
blondinia [14]4 years ago
5 0

Answer:

a quantity expressing the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow

Explanation:

hope this helps

please like and mark as brainliest

matrenka [14]4 years ago
3 0

Answer:

Viscosity Definition Viscosity refers to the thickness of fluid. Viscosity results from the interaction, or friction, between molecules in a fluid. ... In physics, viscosity is often expressed using Isaac Newton's equation for fluids, which is similar to Newton's second law of motion.

HOPE IT HELPS YOU !!!!!!!!!

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What are the names of the ions?
aivan3 [116]

Answer:

Well for most of them you add -ide

Explanation:

4 0
3 years ago
A baseball player throws a ball horizontally at the same moment another baseball player jobs a second ball straight down from th
attashe74 [19]

I don't know what you mean when you say he "jobs" the other ball, and the answer to this question really depends on that word.

I'm going to say that the second player is holding the second ball, and he just opens his fingers and lets the ball <u><em>drop</em></u>, at the same time and from the same height as the first ball.

Now I'll go ahead and answer the question that I've just invented:

Strange as it may seem, <em>both</em> balls hit the ground at the <em>same time</em> ... the one that's thrown AND the one that's dropped.  The horizontal speed of the thrown ball has no effect on its vertical acceleration, so both balls experience the same vertical behavior.

And here's another example of the exact same thing:

Say you shoot a bullet straight out of a horizontal rifle barrel, AND somebody else <em>drops</em> another bullet at exactly the same time, from a point right next to the end of the rifle barrel.  I know this is hard to believe, but both of those bullets hit the ground at the same time too, just like the baseballs ... the bullet that's shot out of the rifle and the one that's dropped from the end of the barrel.

7 0
3 years ago
Help me plz it physics​
lesantik [10]
With what you do need help?????
7 0
4 years ago
Now, let's see what happens when the cannon is high above the ground. Click on the cannon, and drag it upward as far as it goes
kipiarov [429]

Answer:

<h2>B. 20°</h2>

Explanation:

Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g

U is the initial velocity of the body (in m/s)

Ф is the angle of projection

g is the acceleration due to gravity.

Given U = 14m/s, g = 9.8m/s and range R = 15 m

we will substitute this value into the formula to get the projection angle Ф as shown;

15 = 15²sin2Ф/9.8

15*9.8 = 15²sin2Ф

147 = 225sin2Ф

sin2Ф = 147/225

sin2Ф = 0.6533

2Ф = sin⁻¹0.6533

2Ф = 40.79°

Ф = 40.79°/2

Ф = 20.39° ≈ 20°

Hence, the range is greatest at angle 20°

5 0
3 years ago
The velocity of a 640-kg auto is changed from 10.0 m/s to 44.0 m/s in 71.0 s by an external, constant force. (a) What is the res
n200080 [17]

Answer:

(a) <em>change in momentum of the car = 21760 kgm/s</em>

(b) <em>The magnitude of the force = 306.48 N</em>

Explanation:

<em>(a)</em>

<em>Change in momentum: </em><em>This is the product of mass and change in velocity. The unit of momentum is kgm/s. It can be expressed mathematically as,</em>

Change in momentum of the car = m(v-u)................. Equation 1

Where m = mass of the car, v = final velocity of the car, u = initial velocity of the car.

<em>Given: m = 640 kg, v = 44.0 m/s, u = 10.0 m/s.</em>

<em>Substituting these values into equation 1,</em>

<em>Change in momentum of the car = 640 (44-10)</em>

<em>change in momentum of the car = 640× 34</em>

<em>change in momentum of the car = 21760 kgm/s</em>

<em>change in momentum of the car = 21760 kgm/s</em>

<em>(b) Force:</em><em> Force of a body can be defined as the product of mass and its acceleration. It is measured in Newton (N). It can be expressed mathematically as,</em>

<em>Force = ma ................................ Equation 2</em>

<em> </em><em>    Or</em>

<em>Force = m(v-u)/t...................................... Equation 3</em>

<em>Where m = mass of the car, v = final velocity of the car, u = initial velocity of the car, a = acceleration of the car, t = time taken to change the velocity.</em>

<em>Given: m = 640 kg, u = 10.0 m/s v = 44 m/s, t = 71 s.</em>

<em>Substituting these values into equation 3</em>

<em>Force = 640(44-10)/71</em>

<em>Force = (640×34)/71</em>

<em>Force = 306.48 N</em>

<em>The magnitude of the force = 306.48 N</em>

3 0
3 years ago
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