Answer:
5m/s
Explanation:
p=mv, or momentum (p) is equal to mass (m) times velocity (v).
so:
m=5Kg
p=25Kgm/s
v=p÷m
v=25÷5
v=5m/s
hoped this helped :)
Answer:
10.95 minute
Explanation:
V = 120 V, I = 2 A, m = 0.489 kg, c = 4186 J/kgC, T1 = 23 C , T2 = 100 C
Let time be the t.
Heat energy is equal to electrical energy
m x c (T2 - T1) = V x I x t
0.489 x 4186 x (100 - 23) = 120 x 2 x t
t = 656.73 second
t = 10.95 minute
Answer:
Area of the rectangular field in kilometers is 0.09 ![km^2](https://tex.z-dn.net/?f=km%5E2)
Explanation:
We know that 1 kilometers = 1000 meters
since we need to find the area in unit of kilometers
therefore converting length and width into kilometers
1000 meters = 1 kilometers
300 meters =![\frac{300}{1000} = 0.3](https://tex.z-dn.net/?f=%5Cfrac%7B300%7D%7B1000%7D%20%3D%200.3)
Likewise width = 0.3 km
Area = length x width
= 0.3km x 0.3 km
= 0.09 ![km^2](https://tex.z-dn.net/?f=km%5E2)
Answer:
a) 2250 J
b) 0 J
c) 2250 J
Explanation:
a) Since, the process is isochoric
the change in internal energy
![\Delta U = n C_v(T_f-T_i)](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20n%20C_v%28T_f-T_i%29)
Here, n = 0.2 moles
Cv = 12.5 J/mole.K
We have to find T_f so we can use gas equation as
![\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BP_2V_2%7D%20%3D%5Cfrac%7BT_i%7D%7BT_f%7D%5C%5CSince%2C%20V_1%3DV_2%20%20%20%20%5Bisochoric%2Fprocess%5D%5C%5C%5CRightarrow%20%5Cfrac%7BP_%7Batm%7D%7D%7B4P_%7Batm%7D%7D%20%3D%20%5Cfrac%7B300%7D%7BT_f%7D%20%5C%5C%5CRightarrow%20T_f%20%3D%201200%20K)
So, ![\Delta U= 0.2\times12.5(1200-300)\\=2250 J](https://tex.z-dn.net/?f=%5CDelta%20U%3D%200.2%5Ctimes12.5%281200-300%29%5C%5C%3D2250%20J)
b) Since, the process is isochoric no work shall be done.
c) By first law of thermodynamics we have
![\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20Q-W%5C%5CSince%2C%20W%20%3D%200%5C%5C%5CDelta%20U%20%3D%20Q%5C%5CTherefore%2C%20Q%20%3D%202250%20J)
Since, Q is positive 2250 J of heat will flow into the system.
Answer:
a)
16.33 Ω
b)
45009.18 A
Explanation:
a)
L = length of the line = 935 km = 935000 m
d = diameter of the line = 3.50 cm = 0.035 m
ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m
Area of cross-section of the line is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.035)²
A = 0.000961625 m²
Resistance of the line is given as
![R=\frac{\rho L}{A}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Crho%20L%7D%7BA%7D)
inserting the values
R = (1.68 x 10⁻⁸) (935000)/(0.000961625)
R = 16.33 Ω
b)
V = potential difference across the line = 735 kv = 735000 Volts
i = current carried by the wire
Using ohm's law, current carried by the wire is given as
![i=\frac{V}{R}](https://tex.z-dn.net/?f=i%3D%5Cfrac%7BV%7D%7BR%7D)
i = 735000/16.33
i = 45009.18 A