Answer:
0.68 m
Explanation:
α = dL / L1*(dT)
dL = L1(dT) * α
Initial length, L1 = 100
Chang in Length =dL
α linear expansivity ; dL = change in length ; dT = change in temperature ; L1 = initial length
α of iron rod = 1.13 * 10^-5 k
dL = 100(40 - 10) * 1.13 * 10^-5
dL = 100(30) * 1.13 * 10^-5
dL = 3000 * 1.13 * 10^-5
dL = 3390 * 10^-5
dL = 0.0339 m
Error :
Distance measured = 2km = (2 * 1000) = 2000m
[Distance measured / (initial length + change in length)] × change in length
Error = (2000 / (100 + 0.0339)) * 0.0339
Error = (2000 / 100.0339) * 0.0339
Error = 19.993222 * 0.0339
Error = 0.6777702
Error = 0.68 m
1) Yes
2) 
Explanation:
1)
To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.
The pressure exerted by a column of fluid of height h is:
where
is the atmospheric pressure
is the fluid density
is the acceleration due to gravity
h is the height of the column of fluid
Here we have:
is the sea water density
h = 5440 m is the depth at which the bathyscaphe is located
Therefore, the pressure on it is
Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.
2)
Here we want to find the force exerted on the bathyscaphe.
The relationship between force and pressure on a surface is:
where
p is hte pressure
F is the force
A is the area of the surface
Here we have:
is the pressure exerted
The bathyscaphe has a spherical surface of radius
r = 3 m
So its surface is:
Therefore, we can find the force exerted on it by re-arranging the previous equation:
Answer:
the internal energy of the gas is 433089.52 J
Explanation:
let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.
the internal energy of an ideal gas is given by:
Ein = 3/2×n×R×T
= 3/2×(5.3)×(8.31451)×(24 + 273)
= 433089.52 J
Therefore, the internal energy of this gas is 433089.52 J.
Answer:
Option A is correct.
when it is used in a circuit. its terminal voltage will be less than 1.5 V.
Explanation:
The terminal voltage of the battery when it is in use in circuits drops lower than the 1.5 V rating given to it due to internal resistance.
All batteries give internal resistances when used in circuits. The internal resistance (though very small) is usually modelled as connected in series with the battery. It is due to some form of interference from the chemical makeup of the battery.
Normally, while the battery is fresh, the voltage (V) obtained at its terminals when connected in series with a resistor of resistance R is V = IR; where I is the current flowing in this circuit.
But once the interenal resistance (r) of the battery comes into play,
V = I₁ (r + R)
The current in the circuit evidently drops (that is I₁ < I) and V = (I₁r + I₁R)
The voltage across the terminals of the battery is no longer V but is now (V) × [R/(R+r)] which is less than the initial V and it reduces as the internal resistance, r, increases.
Hope this Helps!!!
