Answer:
Stephen hawking if his family were scientists
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
Read more about vertical weight here:
brainly.com/question/15244771
#SPJ1
Answer:
1.69 T
Explanation:
Applying,
F = BvqsinФ.................. Equation 1
Where F = Force, B = magnetic field, v = velocity, q = charge on an electron, Ф = angle between the electron and the field.
make B the subject of the equation,
B = F/(vqsinФ)............. Equation 2
From the question,
Given: F = 2.0×10⁻¹³ N, v = 7.4×10⁵ m/s, Ф = 90°
Constant: q = 1.60×10⁻¹⁹ C
Substitute into equation 2
B = 2.0×10⁻¹³/(7.4×10⁵×1.60×10⁻¹⁹×sin90°)
B = 0.169×10
B = 1.69 T
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J