The answer is bend towards normal.
Answer:
13.4 x 10 raise to power -19 C
Explanation:
. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m
. The uniform electric field is E = 214 N/M
, The decrease in electrical potential energy is
d(P.E) = 51.63 x 10 raise to power -19 J
Let the magnitude of the charge of the moving particle be q
which is given by the equation
d(P.E) =qEd
51.63 x 10 power -19 = q(214)(0.018)
51.63 x 10 power -19 =3.852q
by making q the formular,
q = 13.4 x 10 power -19 C
Answer:
a = 1.16 m/s²
Explanation:
In order to find the acceleration of the ball we will use 3rd equation of motion.
2as = Vf² - Vi²
where,
a = acceleration = ?
s = displacement = 21.9 m
Vf = Final Velocity = 7.14 m/s
Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)
Therefore, using the values, we get:
2a(21.9 m) = (7.14 m/s)² - (0 m/s)²
a = (50.97 m²/s²)/(43.8 m)
<u>a = 1.16 m/s²</u>
D. Reflects
The object absorbs the rest of the color spectrum
