To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.
Fg = m • g
Fg = 1.3 kg • 9.8 m/s^2
Fg = 12.74 N or about 12.7 N.
The solution is C. 12.7 N.
Answer: 1.76 Nm
Explanation:
If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.
In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:
τ = F * r = 16 N. (0.11) m = 1.76 Nm
Answer:
Fₓ = 0, 
= 0 and 
<em> = - 3.115 10⁻¹⁵ N</em>
Explanation:
The magnetic force given by the expression
F = q v xB
the bold are vectors, the easiest analytical way to determine this force in solving the determinant
![F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}](https://tex.z-dn.net/?f=F%20%3D%20q%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C5.8%26-6.7%260%5C%5C-0.81%260.6%260%5Cend%7Barray%7D%5Cright%5D%20%2010%5E%7B4%7D)
F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]
F =i^0 + j^0
- k^ 3.115 10⁻¹⁵ N
Fₓ = 0
= 0
<em> = - 3.115 10⁻¹⁵ N</em>
Soory i think multiple of velocity and angle
Answer:
the Voltages become negative instead of positive but the magnitude remains the same.
Explanation:
If we remove the positive charge by dragging it back to the box at the bottom, and drag a negative charge (blue) toward the middle of the screen, the Voltages become negative instead of positive but the magnitude remains the same.
