Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
Below are the choices that can be found elsewhere:
A. (4.9 × 10-14 newtons) · tan(30°)
<span>B. (4.9 × 10-14 newtons) · sin(30°) </span>
<span>C. (4.9 × 10-14 newtons) · cos(30°) </span>
<span>D. (4.9 × 10-14 newtons) · arctan(30°) </span>
<span>E. (4.9 × 10-14 newtons) · arccos(30°)
</span>
<span>Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)</span>
Answer: A PN-junction diode is formed when a p-type semiconductor is fused to an n-type semiconductor creating a potential barrier voltage across the diode junction
Explanation: hope this helped
Answer:
400000
Explanation:
So first solve one part:
(3.25 * 10^5)
(3.25 * 100,000)
= 325000
Then solve the next part:
(7.5 * 10^4)
(7.5 * 10000)
= 75000
Now lastly, add the two answers:
325000 + 75000 = 400000
Therefore,
(3.25 x 10^5) + (7.5 x 10^4) = 400000
Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
<span>Wt. of skier. </span>
<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>
<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>
<span>Ek + Ep = Ekmax - Fk*d </span>
<span>Ek = Ekmax-Ep-Fk*d </span>
<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>