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slavikrds [6]
4 years ago
11

UPVOTES FOR EVERY ANSWER!!!!!!!!

Physics
2 answers:
r-ruslan [8.4K]4 years ago
6 0

Answer:

Option (c) is correct.

Explanation:

There are some properties of magnetic field lines which are given below:

(i) The field lines begin near the magnetic north pole and extend towards the south pole of the magnet outside the magnet.

(ii) Inside the magnet, the magnetic filed lines moves from south pole to north pole.

(iii) The tangent at any point on the field line gives the direction of magnetic field at that point.

mario62 [17]4 years ago
4 0
<span>C. Field lines begin near the magnet’s north pole and extend toward its south pole. 

Hope this helps!</span>
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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

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As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

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4 years ago
2.)
Oduvanchick [21]

Answer:

-22.7 m/s^2

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This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

v^2-u^2=2as

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u = 27.8 m/s

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s = 17 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2

4 0
3 years ago
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