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Ugo [173]
3 years ago
14

The force that pushes against something traveling through a fluid is called

Physics
1 answer:
olganol [36]3 years ago
3 0

Viscous force is the force that pushes against something travelling through a fluid. Viscous force is an analogue in fluids(liquids or gases) of the force of friction. It acts opposite to the direction in which the object is moving relative to the fluid.

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In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.

Explanation:

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When a current is passed through plasma and that current is increased will the temperature of the plasma increase . if it does t
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The temperature of the plasma will increase as current is passed through it.

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2 years ago
In this section of a circuit, a current of 2.5 A flows across R2. Find the current that flows across R3. Let R1= 3.0 ohm, R2= 8.
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3 years ago
Read 2 more answers
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
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