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Answer:
The apparent weight of the object is 0.465 N.
Explanation:
Given that,
Weight = 0.71 N
Water level = 50 mL
object inserted = 75 mL
We need to calculate the volume of solid
Using formula of volume
![V=25\ ml = 25\times10^{-6}\ m^3](https://tex.z-dn.net/?f=V%3D25%5C%20ml%20%3D%2025%5Ctimes10%5E%7B-6%7D%5C%20m%5E3)
We need to calculate the buoyancy force
Using formula of buoyancy force
![F'= V\rho g](https://tex.z-dn.net/?f=F%27%3D%20V%5Crho%20g)
Put the value into the formula
![F'=25\times10^{-6}\times1000\times9.8](https://tex.z-dn.net/?f=F%27%3D25%5Ctimes10%5E%7B-6%7D%5Ctimes1000%5Ctimes9.8)
![F'=0.245\ N](https://tex.z-dn.net/?f=F%27%3D0.245%5C%20N)
We need to calculate the apparent weight of the object
Using formula of apparent weight
![W=F-F'](https://tex.z-dn.net/?f=W%3DF-F%27)
Put the value into the formula
![W=0.71-0.245](https://tex.z-dn.net/?f=W%3D0.71-0.245)
![W=0.465\ N](https://tex.z-dn.net/?f=W%3D0.465%5C%20N)
Hence, The apparent weight of the object is 0.465 N.
Answer:
Explanation:
To calculate the momentum of a moving object multiply the mass of the object times its velocity. The symbol for momentum is a small p. So, the momentum of the object is calculated to be 8.0 kg-m/s. Note the unit for momentum.
Answer:
a. 11.5kv/m
b.102nC/m^2
c.3.363pF
d. 77.3pC
Explanation:
Data given
![area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v](https://tex.z-dn.net/?f=area%3D7.60cm%5E%7B2%7D%5C%5C%20distance%2Cd%3D2mm%5C%5Cvoltage%2Cv%3D23v)
to calculate the electric field, we use the equation below
V=Ed
where v=voltage, d= distance and E=electric field.
Hence we have
![E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m](https://tex.z-dn.net/?f=E%3Dv%2Fd%5C%5CE%3D%5Cfrac%7B23%7D%7B2%2A10%5E%7B-3%7D%7D%20%5C%5CE%3D11.5%2A10%5E%7B3%7D%20v%2Fm%5C%5CE%3D11.5Kv%2Fm)
b.the expression for the charge density is expressed as
σ=ξE
where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2
If we insert the values we have
![8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}](https://tex.z-dn.net/?f=8.85%2A10%5E%7B-12%7D%20%2A11500%5C%5C1.02%2A10%5E%7B-7%7DC%2Fm%5E%7B2%7D%20%20%5C%5C102nC%2Fm%5E%7B2%7D)
c.
from the expression for the capacitance
![C=eA/d](https://tex.z-dn.net/?f=C%3DeA%2Fd)
if we substitute values we arrive at
![C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B8.85%2A10%5E%7B-12%7D%2A7.6%2A10%5E%7B-4%7D%7D%7B2%2A10%5E%7B-3%7D%20%7D%20%5C%5CC%3D%5Cfrac%7B6.726%2A10%5E%7B-15%7D%20%7D%7B2%2A10%5E%7B-3%7D%20%7D%20%5C%5CC%3D3.363%2A10%5E%7B-12%7DF%5C%5CC%3D3.363pF)
d. To calculate the charge on each plate, we use the formula below
![Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC](https://tex.z-dn.net/?f=Q%3DCV%5C%5CQ%3D23%2A3.363%2A10%5E%7B-12%7D%5C%5C%20Q%3D7.73%2A10%5E%7B-12%7D%5C%5C%20Q%3D77.3pC)