It will be unaffected by the magnet because it has no magnetic field. If you were to maybe have electricity going through it is the only way it would have anything to do with the magnet.
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I think that the answe is gasses.
Answer:
The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.
By definition, the momentum is given by:
p = m * v
Where,
m = mass
v = speed.
On the other hand,
F = m * a
Where,
m = mass
a = acceleration:
For the boy we have:
p1 = m * v
p1 = (F / a) * v
p1 = ((710) / (9.81)) * (0.50)
p1 = 36.19 Kg * (m / s)
For the girl we have:
p2 = m * v
p2 = (F / a) * v
p2 = ((480) / (9.81)) * (v)
p2 = 48.93 * v Kg * (m / s)
Then, we have:
p1 + p2 = 0
36.19 + 48.93 * v = 0
Clearing v:
v = - (36.19) / (48.93)
v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
Answer:
the girl's velocity in m / s after they push off is -0.74 m / s
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
