Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.


Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
Impulse=force x time
force=mass x acceleration due to gravity
force=

impulse =3000 x 2.5= ( sorry i don't have a calculator right now so you must calculate this yourself)
I converted from kg to g because it is the standard.
Hope this helps you.
B) law of conservation of momentum
It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.
In this case total momentum before impact:
10kg*5m/s + 5kg * 0m/s = 50 kg m/s
After Impact:
10kg*0m/s + 5kg*10m/s = 50 kg m/s
You can see the momentum before and after impact is same as 50 kg m/s
Of course we assumed that the first cart stopped after the impact, and there are no energy losses.
Answer:
x = 7.14 meters
Explanation:
It is given that,
Current in wire 1, 
Current in wire 2,
Distance between parallel wires, r = 25 cm
Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

Let the distance is x from wire 1. So,



x = 7.14 meters
So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.
Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

Taking natural log on both sides,

(1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
= 
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns