<h2>
Density of the unknown liquid is 771.93 kg/m³</h2>
Explanation:
An empty graduated cylinder weighs 55.26 g
Weight of empty cylinder = 55.26 g = 0.05526 kg
Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³
Weight of cylinder plus liquid = 92.39 g = 0.09239 kg
Weight of liquid = 0.09239 - 0.05526
Weight of liquid = 0.03713 kg
We have
Mass = Volume x Density
0.03713 = 48.1 x 10⁻⁶ x Density
Density = 771.93 kg/m³
Density of the unknown liquid is 771.93 kg/m³
Answer:
<em>Gravity</em><em>.</em><em> </em><em>The</em><em> </em><em>weight-force</em><em> </em><em>or</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>because</em><em> </em><em>of</em><em> </em><em>Gravity</em><em>,</em><em> </em><em>which</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>object</em><em> </em><em>attracting</em><em> </em><em>it</em><em> </em><em>towards</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>earth</em><em>.</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>
If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m
Answer:
3.98V
Explanation:
Given
Pontential difference V as 3v
Energy delivered is 30%,
Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)
So E=V^2
E=3^2=9
At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3
But E=V^2
✓E=V
✓6.3=3.98V
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