The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is
<em>F</em> = <em>G M m</em> / <em>r</em> ²
Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is
<em>F*</em> = <em>G </em>(<em>M</em> / 8) (<em>m</em> / 8) / (<em>r</em> / 7)²
<em>F*</em> = (1/64 × <em>G M m</em>) / (1/49 × <em>r</em> ²)
<em>F*</em> = 49/64 × <em>G M m</em> / <em>r</em> ²
In other words, the new force is scaled down by a factor of 49/64 ≈ 0.7656, so the new force has magnitude approx. 76.56 N.
Answer:
Explanation:
Electric field between plates of a parallel plate capacitor is uniform .
In a uniform electric field , relation between electric field and potential gradient is as follows
electric field = potential gradient [ E = - dV / dl ]
in the given case ,
dV = 51 V ,
dl = 4 cm
= 4 x 10⁻² m
E = 51 / 4 x 10⁻²
= 12.75 x 10² V / m
= 1275 V / m
Answer:
I think it's 2 the photo is hard to tell what they are exactly talking about.