You can use the displacement method or the eureka can so basically in the displacement can what you have to do is to put some water into a measuring cylinder and measure its volume before adding the irregular shaped object and then measuring the level of water which had been displaced and then eureka can you can check online
Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
Answer:molecules slow down
Explanation:
because science
Answer:
See the answers below.
Explanation:
The cost of energy can be calculated by multiplying each given value, a dimensional analysis must be taken into account in order to calculate the total value of the cost in Rs.
![Cost=0.350[kW]*12[\frac{hr}{1day}]*30[days]*4.5[\frac{Rs}{kW*hr} ]=567[Rs]](https://tex.z-dn.net/?f=Cost%3D0.350%5BkW%5D%2A12%5B%5Cfrac%7Bhr%7D%7B1day%7D%5D%2A30%5Bdays%5D%2A4.5%5B%5Cfrac%7BRs%7D%7BkW%2Ahr%7D%20%5D%3D567%5BRs%5D)
The fuse can be calculated by knowing the amperage.
where:
P = power = 350 [W]
V = voltage = 240 [V]
I = amperage [amp]
Now clearing I from the equation above:
![I=P/V\\I=350/240\\I=1.458[amp]](https://tex.z-dn.net/?f=I%3DP%2FV%5C%5CI%3D350%2F240%5C%5CI%3D1.458%5Bamp%5D)
The fuse should be larger than the current of the circuit, i.e. about 2 [amp]
