Answer : The correct option is, (c) ![3.7\times 10^2J/^oC](https://tex.z-dn.net/?f=3.7%5Ctimes%2010%5E2J%2F%5EoC)
Explanation :
First we have to calculate the energy or heat.
Formula used :
![E=V\times I\times t](https://tex.z-dn.net/?f=E%3DV%5Ctimes%20I%5Ctimes%20t)
where,
E = energy (in joules)
V = voltage (in volt)
I = current (in ampere)
t = time (in seconds)
Now put all the given values in the above formula, we get:
![E=(3.6V)\times (2.6A)\times (350s)](https://tex.z-dn.net/?f=E%3D%283.6V%29%5Ctimes%20%282.6A%29%5Ctimes%20%28350s%29)
![E=3276J](https://tex.z-dn.net/?f=E%3D3276J)
Now we have to calculate the heat capacity of the calorimeter.
Formula used :
![C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7BE%7D%7B%5CDelta%20T%7D%3D%5Cfrac%7BE%7D%7BT_%7Bfinal%7D-T_%7Binitial%7D%7D)
where,
C = heat capacity of the calorimeter
= initial temperature = ![20.3^oC](https://tex.z-dn.net/?f=20.3%5EoC)
= final temperature = ![29.1^oC](https://tex.z-dn.net/?f=29.1%5EoC)
Now put all the given values in this formula, we get:
![C=\frac{3276J}{(29.1-20.3)^oC}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B3276J%7D%7B%2829.1-20.3%29%5EoC%7D)
![C=372.27J/^oC=3.7\times 10^2J/^oC](https://tex.z-dn.net/?f=C%3D372.27J%2F%5EoC%3D3.7%5Ctimes%2010%5E2J%2F%5EoC)
Therefore, the heat capacity of the calorimeter is, ![3.7\times 10^2J/^oC](https://tex.z-dn.net/?f=3.7%5Ctimes%2010%5E2J%2F%5EoC)
Theres: the vacuole, nucleus, rough endoplamid reticulum, smooth endoplasmic reticulum, cell memebrane, cell wall, chloroplast, mitochondria, golgi apperatus, lysosomes, and ribosomes
relation between potential difference and electric field is given as
![E . d = \Delta V](https://tex.z-dn.net/?f=E%20.%20d%20%3D%20%5CDelta%20V)
so here we know that
d = 3 cm
![\Delta V = 30 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%2030%20V)
![E \times 0.03 = 30](https://tex.z-dn.net/?f=E%20%5Ctimes%200.03%20%3D%2030)
![E = 1000 N/C](https://tex.z-dn.net/?f=E%20%3D%201000%20N%2FC)
So now when plates are separated to 4 cm distance carefully
the potential difference between them will change but the electric field between them will remain constant
So at distance of 4 cm also the electric field will be E = 1000 N/C
Answer:
Wn = 9.14 x 10¹⁷ N
Explanation:
First we need to find our mass. For this purpose we use the following formula:
W = mg
m = W/g
where,
W = Weight = 675 N
g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²
m = Mass = ?
Therefore,
m = (675 N)/(9.8 m/s²)
m = 68.88 kg
Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:
gn = (G)(Mn)/(Rn)²
where,
gn = acceleration due to gravity on surface of neutron star = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg
Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m
Therefore,
gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)
gn = 13.27 x 10¹⁵ m/s²
Now, my weight on neutron star will be:
Wn = m(gn)
Wn = (68.88)(13.27 x 10¹⁵ m/s²)
<u>Wn = 9.14 x 10¹⁷ N</u>
Answer:
Given: V = 220V, Pmin = 360W, Pmax = 840W
For minimum heating case:
We know that
Pmin = VI
360 = 220 X I
I = 1.63 amp
R = V/I
R = 220/1.63
R = 134.96ohms
For maximum heating case:
We know that
Pmax = VI
840 = 220 X I
I = 3.81 amp
R = V/I
R = 220/3.81
R = 57.74 ohms