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>
Answer:
The Kinetic Energy decreases. The Total Energy stays the same
Explanation:
The TE stays the same, so if PE increases then KE will decrease.
1) <span>yes;2
6*2=12
12*2=24
24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1
</span>
a = first term, i.e. 5
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
<span>T4 = 135
</span>T5= ar^n-1
<span>= 5*3^5-1 </span>
<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>
Answer:
I believe the answer is B.
It is transferred evenly to water through air
