Answer:

Explanation:
The total force on the particle is given by

Then, by replacing we have:
![q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N](https://tex.z-dn.net/?f=q%5Cvec%7Bv%7D%5C%20X%20%5Cvec%7BB%7D%3Dq%5B7%5Chat%7Bk%7D-9%5Chat%7Bj%7D-%5Chat%7Bk%7D%5D%5C%5C%5C%5Cq%5Cvec%7BE%7D%3Dq%5B5%5Chat%7Bi%7D-%5Chat%7Bj%7D-2%5Chat%7Bk%7D%5D%5C%5C%5C%5C%5Cvec%7BF%7D%3D%289.61%2A10%5E%7B-19%7DC%29%5B%287%2B9%29%5Chat%7Bi%7D%2B%28-9-1%29%5Chat%7Bj%7D%2B%28-1-2%29%5Chat%7Bk%7D%5D%5C%5C%5C%5C%5Cvec%7BF%7D%3D%281.537%2A10%5E%7B-17%7D%5Chat%7Bi%7D-9.61%2A10%5E%7B-19%7D%5Chat%7Bj%7D-2.883%2A10%5E%7B-18%7D%5Chat%7Bk%7D%29N)
where the cross product can be made with the determinant method.
Hope this helps!!
Answer:
cargo planes hold cargo so there hevier
Explanation:
The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
<h3>
Acceleration of the cart</h3>
The acceleration of the cart is determined from the net force acting on the mass-cart system.
Upward force = Downward force
ma = mg
13a = 10(9.8)
13a = 98
a = 98/13
a = 7.54 m/s²
Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
Learn more about acceleration here: brainly.com/question/14344386
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Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and

h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that


So on

= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx