Answer:
The magnitude and direction of the force between the two charges is 497.66 N in opposite direction.
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 1.4×10¯³ C
Charge 2 (q₂) = 9.1×20¯⁴ C
Distance apart (r) = 4.8 m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The magnitude of the force between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 1.4×10¯³ × 9.1×20¯⁴ / 4.8²
F = 11466 / 23.04
F = 497.66 N
Thus, the magnitude of the force is 497.66 N.
Since both charges has the same sign (i.e they both positive), the force between them will be a force of repulsion. Hence the force will be in opposite direction.
Answer:
110 m/s²
Explanation:
Given:
v₀ = 0 m/s
Δx = 1.08 m
t = 0.14 s
Find: a
Δx = v₀ t + ½ at²
1.08 m = (0 m/s) (0.14 s) + ½ a (0.14 s)²
a = 110 m/s²
Answer:
4611.58 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32.174 ft/s²
Equation of motion
Magnitude of acceleration while stopping is 4611.58 ft/s²
Answer:
TRUE
Explanation:
Protons have a positive charge. Electrons have a negative charge. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.
Answer:
e. The torque is the same for all cases.
Explanation:
The formula for torque is:
τ = Fr
where,
τ = Torque
F = Force = Weight (in this case) = mg
r = perpendicular distance between force an axis of rotation
Therefore,
τ = mgr
a)
Here,
m = 200 kg
r = 2.5 m
Therefore,
τ = (200 kg)(9.8 m/s²)(2.5 m)
<u>τ = 4900 N.m</u>
<u></u>
b)
Here,
m = 20 kg
r = 25 m
Therefore,
τ = (20 kg)(9.8 m/s²)(25 m)
<u>τ = 4900 N.m</u>
<u></u>
c)
Here,
m = 8 kg
r = 62.5 m
Therefore,
τ = (8 kg)(9.8 m/s²)(62.5 m)
<u>τ = 4900 N.m</u>
<u></u>
Hence, the correct answer will be:
<u>e. The torque is the same for all cases.</u>
