Answer:
The charge carried by the droplet is 
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude 
points straight down and if the mass of the droplet is 
We need to calculate the acceleration
Using equation of motion
Put the value into the formula
We need to calculate the charge carried by the droplet
Using formula of electric filed
Put the value into the formula
Hence, The charge carried by the droplet is
The formula for the potential energy, EP is given by
EP = m × g × h
where:
m is the mass in kilograms
g is the gravity (9.8 m/s²)
h is the height in meters
EP = 6 × 9.8 × 2 = 117.6 kg-m²/s²
Answer:
0.06 Kg
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Net Force (F) = 3 N
Mass (m) =?
Next, we shall determine the acceleration of the object. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 3.0 m/s
Distance (s) = 0.09 m
Acceleration (a) =?
v² = u² + 2as
3² = 0² + (2 × a × 0.09)
9 = 0 + 0.18a
9 = 0.18a
Divide both side by 0.18
a = 9 / 0.18
a = 50 m/s²
Finally, we shall determine the mass of the object. This can be obtained as follow:
Net Force (F) = 3 N
Acceleration (a) = 50 N
Mass (m) =?
F = ma
3 = m × 50
Divide both side by 50
m = 3 / 50
m = 0.06 Kg
Therefore, the mass of the object is 0.06 Kg
Answer:
1.)The field energy will increase.
The rest of the answers:
2.)The energy increases, and the lines of force are denser
3.) It points toward the field of earths magnetic poles.
4.) l, ll, and lll only
5.) ll, lV, l, lll
