This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
<em>Hint </em>: ΔV = Ed <em>
</em>
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F
= 0
so, F
= F
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
Answer:
Secondary structure
The secondary structure arises from the hydrogen bonds formed between atoms of the polypeptide backbone. The hydrogen bonds form between the partially negative oxygen atom and the partially positive nitrogen atom
Answer:
ooh thanks but don't give me advise
By gravity as well as electrostatic and magnetic attraction and repulsion
Example forces being exerted by one object on another without them being in contact with each other
Answer:
u = - 20 cm
m =
Given:
Radius of curvature, R = 10 cm
image distance, v = 4 cm
Solution:
Focal length of the convex mirror, f:
f = 
Using Lens' maker formula:

Substitute the given values in the above formula:


u = - 20 cm
where
u = object distance
Now, magnification is the ratio of image distance to the object distance:
magnification, m =
magnification, m =
m =
m =