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:
<em>voltage</em><em> </em><em>,</em><em>current</em><em> </em><em>and</em><em> </em><em>resistance</em><em> </em>
Answer:
4987N
Explanation:
Step 1:
Data obtained from the question include:
Mass (m) = 0.140 kg
Initial velocity (U) = 28.9 m/s
Time (t) = 1.85 ms = 1.85x10^-3s
Final velocity (V) = 37.0 m/s
Force (F) =?
Step 2:
Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:
F = m(V + U) /t
F = 0.140(37+ 28.9) /1.85x10^-3
F = 9.226/1.85x10^-3
F = 4987N
Therefore, the magnitude of the horizontal force applied is 4987N
Answer:
The horizontal distance traveled by the projectile is 60 m
Explanation:
Given;
initial horizontal velocity of the projectile, Vₓ = 30 m/s
time of the motion of the projectile, t = 2 s
The horizontal distance traveled by the projectile is given by the range of the projection;
X = Vₓt
X = 30 x 2
X = 60 m
Therefore, the horizontal distance traveled by the projectile is 60 m
Therefoe
Answer:
A
Contact forces are forces that require the actual contact (touching) of two pieces of matter. ... A field force is a force that works at a distance. No touching is required. Gravity is a good example of a field force, because it works whether or not an object is touching something or touching nothing at all.
