Answer:
4.3 * 10 N
Explanation:
To calculate torque, we multiply the distance from the pivot by the perpendicular (the part of the force that acts at right angles to the displacement vector) component of the force to the displacement vector from the pivot.
torque = distance from pivot * perpendicular force
170 Nm= 0.4 m * F
F = 425 N = 4.3 * 10 N rounded off to two significant figures
Power = (voltage) x (current). The motor consumes (12)x(206)=2,472 watts. Some of it is dissipated as heat, but most of it is used to do useful work by turning the engine over to make it start.
Explanation:
<em>Here </em><em>it </em><em>is </em><em>given </em>
<em>Work </em><em>(</em><em>W) </em><em> </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em> </em><em>J</em>
<em>Time </em><em>(</em><em>t) </em><em> </em><em>=</em><em> </em><em>5</em><em>5</em><em> </em><em>sec</em>
<em>power </em><em>(</em><em>P) </em><em> </em><em>=</em><em> </em><em>?</em>
<em>We </em><em>know </em><em>we </em><em>have </em><em>the </em><em>formula </em>
<em>
</em>
<em>P </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em>/</em><em>5</em><em>5</em>
<em>P </em><em>=</em><em> </em><em>6</em><em>4</em><em>.</em><em>7</em><em>3</em><em> </em><em>watt</em>
Answer:
see explanation
Explanation:
Given that,
velocity of 1.50 km/s = 1.50 × 10³m/s
acceleration of 2.00 ✕ 1012 m/s2
electric field has a magnitude of strength of 18.0 N/C
![\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]](https://tex.z-dn.net/?f=%5Cbar%20F%3D%20q%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%5Cbar%20B%5D%5C%5C%5C%5C%5Cbar%20F%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D%5C%5C%5C%5C%5C%5Cm%20%5Cbar%20a%20%3D%20%5B%5Cbar%20E%20%2B%20%5Cbar%20V%20%5Ctimes%20%28%20B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20z%20%29%5D)
![9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)]](https://tex.z-dn.net/?f=9.1%20%5Ctimes%2010%5E-%5E3%5E1%20%5Ctimes%202%5Ctimes%2010%5E1%5E2%20%5Chat%20k%3D-1.6%5Ctimes10%5E-%5E1%5E9%20%5Chat%20k%20%5B18%5Chat%20k%2B%201.5%5Ctimes%2010%5E3%20%5Chat%20i%20%5Ctimes%20%28B_x%20%5Chat%20i%20%2BB_y%20%5Chat%20j%20%2BB_z%20%5Chat%20k%29%5D)



