D I think .... don’t be mad if I’m wrong
Answer:
J for impulse
t for time
F for force
formula is J=F×t
Explanation:
putting values in eqs after rearranging
we need to find force so
F=J ÷t
F=400÷15
=26.67
=27(rounded off)
27N is the Force applied.
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Answer:
For C1, Q = 1.6125×10⁻³ C
For C2, Q = 6.25×10⁻⁴ C
Explanation:
Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.
From the question,
Q = CV........................ Equation 1
Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.
For the first capacitor,
Q = C1V............. Equation 2
Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V
Substitute into equation 2
Q = (6.45×10⁻⁶ )(250)
Q = 1.6125×10⁻³ C.
For the the second capacitor,
Q = C2V............. Equation 3
Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V
Q = (2.5×10⁻⁶ )(250)
Q = 6.25×10⁻⁴ C
Answer:
a)
Weight in Air = 0.3N
Weight in Water = 0.25N
Weight in Liquid = 0.24N.
Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)
= 0.3 – 0.25
= 0.5N.
b) R.D of Body = Density of Body/Density of Standard Fluid(Water).
There's a Derived Formula for RD.
I'm gonna Apply it here.
Ask me for the derivation in the Comment section if you need it.
RD = α/ρ = (Weight in Air) / (Upthrust Force)
Where
α = density of the Body(or reference substance)
ρ = density of standard fluid (water)
= 0.3/0.05 = 6.
c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)
Or we just go by that formula
RD of Liquid = Weight in Air/Upthrust(In Liquid)
We'll be using the Upthrust in that Liquid now.
= 0.3 – 0.24 = 0.06
RD = 0.3/0.06 = 5.