Answer:
Explanation:
Force on the electron = q ( v x B )
q = - 1.6 x 10⁻¹⁹
v = (5.9i−6.4j)×10⁴
B = (−0.63i+0.65j)
v x B = (5.9i−6.4j)×10⁴ x (−0.63i+0.65j)
= (3.835 - 4.032 ) x 10⁴ k
= - 1970 k
Force on the electron = q ( v x B )
= - 1.6 x 10⁻¹⁹ x -1970 k
= 3.152 x 10⁻¹⁶ k
z-component of the force on the electron
Fz = 3.152 x 10⁻¹⁶ N
Acceleration = (change in speed) / (time for the change.
change in speed = (ending speed) - (starting speed) = 9 m/s.
Acceleration = (9 m/s) / (period of 1) .
We don't know the units of the 'period of 1'.
If it means '1 second', then the acceleration is 9 m/s² .
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:

Calculating the value of 
![\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j]](https://tex.z-dn.net/?f=%5Cto%20%5Cleft%20%7C%20%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7BK%7D%5C%5CR_i%26R_j%26R_k%5C%5CS_i%26S_j%26S_k%5Cend%7Barray%7D%5Cright%20%7C%20%3D%20%5Chat%7Bi%7D%5BR_j%20S_k-S_jR_k%5D-%5Chat%7Bj%7D%5BR_i%20S_k-S_iR_k%5D%2B%5Chat%7Bk%7D%5BR_i%20S_j-S_iR_j%5D)
Calculating the value of 
![\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])](https://tex.z-dn.net/?f=%5Cto%20%28R_i%5Chat%7Bi%7D%2BR_j%5Chat%7Bj%7D%2BR_k%5Chat%7Bk%7D%29%20%5Ccdot%20%28%20%5Chat%7Bi%7D%5BR_j%20S_k-S_jR_k%5D-%5Chat%7Bj%7D%5BR_i%20S_k-S_iR_k%5D%2B%5Chat%7Bk%7D%5BR_i%20S_j-S_iR_j%5D%29)
by solving this value it is equal to 0.
Answer:
Yes
Explanation:
The variables change in and experiment.
Gravity on Earth pulls objects to the center of the Earth about 9.8 meters per second per second. The child's mass on the moon or at a space station is 40 kg. The child's weight at these places changes because gravity is different at these locations.