Answer:
Explanation:
According to the first Newton's law, when a particle moves with constant velocity, the sum of forces on it is zero. So, we have:
Here 
is the magnetic force and 
is the gravitational force. The velocity of the particle is perpendicular to the magnetic field, so 
. Replacing and solving for B:
Answer:
D. unsaturated fat
Explanation:
Unsaturated fats are considered the healthiest fats because they improve cholesterol, help reduce inflammation (a risk factor for heart disease), and help decrease the overall risk of developing heart disease. The main source of unsaturated fats are plant-based foods.
The Sun is a huge, glowing sphere of hot gas. Most of this gas is hydrogen (about 70%) and helium (about 28%). Carbon, nitrogen and oxygen make up 1.5% and the other 0.5% is made up of small amounts of many other elements such as neon, iron, silicon, magnesium and sulfur.
So typically I would say hydrogen and helium.
Answer:
d) The trampoline pushes back up on the gymnast.
Explanation:
According to Newton's third law of motion; for every action force, there is an equal and opposite reaction force. The action force and reaction force are reciprocal to one another i.e. they act oppositely to one another. The reaction force acts in an opposite direction to the action force and vice versa.
In this question, a gymnast pushed down on a trampoline during a routine. This is called the ACTION FORCE. In conformity to Newton's third law, the trampoline pushes back up on the gymnast. This opposite force is called the REACTION FORCE.
Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D-sin%28-%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%2Bsin%28%5Cfrac%7B%5Ctheta%7D%7B2%7D%29%5D)
![E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%5Clambda%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
Since 
where;
L = length
Q = charge
λ = density of the charge;
then substituting 
for λ, we have :
![E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%28%5Cfrac%7BQ%7D%7BL%7D%29%7D%7B4%20%5Cpi%20E_oR%7D%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D)
![E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2Q%5Bsin%5Cfrac%7B%5Ctheta%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20E_oLR%7D)
substituting our given parameter; we have:
![E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7B2%286.26%2A10%5E%7B-12%7DC%29%5Bsin%5Cfrac%7B2.59rad%7D%7B2%7D%5D%7D%7B4%20%5Cpi%20%288.85%2A10%5E%7B-12%7DC%5E2%2FN.m%5E2%29%280.235431%29%280.0909%29%7D)
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C
