Answer:
D
Explanation:
descriptive, because scientists are writing down the observations but not making comparisons.
Answer:
The acceleration of the proton is 1.403 x 10⁹ m/s²
Explanation:
Given;
speed of proton, v = 7.7 m/s
magnitude of magnetic field, B = 1.9 T
Magnetic force of moving proton is given by;
F = qvBsinθ
Centripetal force on the moving proton is given by;


where;
q is charge of the proton = 1.602 x 10⁻¹⁹ C
m is mass of proton = 1.67 x 10⁻²⁷ kg

Therefore, the acceleration of the proton is 1.403 x 10⁹ m/s²
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
Physical activity is defined as any bodily movement produced by skeletal muscles that requires energy expenditure.