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grigory [225]
3 years ago
12

Pleaseeeee help!!!!!!!!

Physics
1 answer:
Ainat [17]3 years ago
5 0
I did it last year. I thinks its A
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A viscous fluid is flowing through two horizontal pipes. They have the same length, although the radius of one pipe is three tim
Rudik [331]

The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,

Q_1 = Q_2

The definition of flow is given by

Q =VA

Where

V = Velocity

A = Area

The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.

In this way

Q_A = Q_B

Q_B = 2.94*10^2 m^3/s

7 0
3 years ago
Which combination of elements will form an ionic bond?
Anna007 [38]

Answer:

C

Explanation:

8 0
2 years ago
Select the correct answer.
Murrr4er [49]

Answer:

I will choose the answer A

3 0
3 years ago
The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler
guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

v^{2}=u^{2}+ 2as

620^{2}=0^{2}+ 2\times a \times 0.89

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is  215955.06 m/s^2.

6 0
3 years ago
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

we make F, the subject of formula

F = \frac{m}{2x}( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
2 years ago
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