Answer:
1. <u>F = ma</u> <em>F = 0.2kg * 20m/s² = 4Kg * m/s² =</em> 4N
2. <u>F = ma</u> <em>F - 18Kg * 3m/s² = 54Kg * m/s² =</em> 54N
3. <u>F = ma</u> <em>F = 0.025Kg * 5m/s² =</em> 0.125N
4. <u>F = ma</u> <em>F = 50Kg * 4m/s² =</em> 200N
5. <u>F = ma</u> <em>F = 70Kg * 4m/s² =</em> 280N
6. <u>F = ma</u> <em>F = 9Kg * 9.8m/s² =</em> 88.2N
Explanation:
Hope this helps ! ^^
Answer:
spring compressed is 0.724 m
Explanation:
given data
mass = 1.80 kg
spring constant k = 2 × 10² N/m
initial height = 2.25 m
solution
we know from conservation of energy is
mg(h+x) = 0.5 × k × x² ...................1
here x is compression in spring
so put here value in equation 1 we get
1.8 × 9.8 × (2.25+x) = 0.5 × 2× 10² × x²
solve it we get
x = 0.724344
so spring compressed is 0.724 m
car starts from rest

final speed attained by the car is

acceleration of the car will be

now the time to reach this final speed will be



so it required 1.39 s to reach this final speed
Answer:
The potential difference between the plates increases
Explanation:
As we know that the capacitance of the capacitor is given by:
(1)
where
q = charge
C = capacitance
V = Voltage or Potential Difference
Also, the capacitance of a parallel plate capacitor is given as:
(2)
where

A = Area of the plates
D = Separation distance between the plates
Now, from eqn (1) and (2):

Now, from the above eqn we can say that:
Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.
Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.