Answer:
42.99°
Explanation:
= Kinetic friction force
= Pulling force at angle ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
= Weight of the box = 150 N
Kinetic friction force
![F_h=\muN_h](https://tex.z-dn.net/?f=F_h%3D%5CmuN_h)
Pulling force at angle ![\theta](https://tex.z-dn.net/?f=%5Ctheta)
![F_{\theta}=\muN_{\theta}](https://tex.z-dn.net/?f=F_%7B%5Ctheta%7D%3D%5CmuN_%7B%5Ctheta%7D)
N = Pulling force
According to question
![\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N](https://tex.z-dn.net/?f=%5Cfrac%7BF_h%7D%7BF_%7B%5Ctheta%7D%7D%3D%5Cfrac%7B2%7D%7B1%7D%5C%5C%5CRightarrow%20%5Cfrac%7B%5CmuN_h%7D%7B%5CmuN_%7B%5Ctheta%7D%7D%3D2%5C%5C%5CRightarrow%20%5Cfrac%7BN_h%7D%7BN_%7B%5Ctheta%7D%7D%3D2%5C%5C%5CRightarrow%20N_%7B%5Ctheta%7D%3D%5Cfrac%7BN_h%7D%7B2%7D%5C%5C%5CRightarrow%20N_%7B%5Ctheta%7D%3D%5Cfrac%7B150%7D%7B2%7D%5C%5C%5CRightarrow%20N_%7B%5Ctheta%7D%3D75%5C%20N)
Applying Newton's second law in the vertical direction we get
![N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}](https://tex.z-dn.net/?f=N_h-Nsin%5Ctheta%3DN_%7B%5Ctheta%7D%5C%5C%5CRightarrow%20150-110sin%5Ctheta%3D75%5C%5C%5CRightarrow%20%5Ctheta%3Dsin%5E%7B-1%7D%5Cfrac%7B75%7D%7B110%7D%5C%5C%5CRightarrow%20%5Ctheta%3D42.99%5C%20%5E%7B%5Ccirc%7D)
The angle is 42.99°
Answer:
The answer is "Option A".
Explanation:
In the given scenario, Two metal plates were produced in contrast to values separated by a small distance and three potential points for just a slight net charge are accessible. All three points would have the same electrical energy since the power generation is consistent from both sides as well as the position is placed at a slight net value so it has the highest electrical energy.
Unless the second ball's speed is twice the amount of the first ball's no they will not cross paths
Answer:
15,000 m
Explanation:
![speed = \frac{distance}{time}](https://tex.z-dn.net/?f=speed%20%3D%20%20%5Cfrac%7Bdistance%7D%7Btime%7D%20)
speed = 50 m/s
time = 5 minutes × 60
= 300 seconds
substitute the values into the formula
![50 = \frac{distance}{300}](https://tex.z-dn.net/?f=50%20%3D%20%20%5Cfrac%7Bdistance%7D%7B300%7D%20)
![50 \times 300 = distance](https://tex.z-dn.net/?f=50%20%5Ctimes%20300%20%3D%20distance)
![15000 \: m = distance](https://tex.z-dn.net/?f=15000%20%5C%3A%20m%20%3D%20distance)
![distance = 15000 \: m](https://tex.z-dn.net/?f=distance%20%20%3D%2015000%20%5C%3A%20m)