Answer:
E(final)/E(initial)=2
Explanation:
Applying the law of gauss to two parallel plates with charge density equal σ:
So, if the charge is doubled the Electric field is doubled too
E(final)/E(initial)=2
Answer:
7.89 m/s
Explanation:
Given that
Width of the river, b1 = 12 m
Width of the river, b2 = 5.8 m
Depth of the river, d1 = 2.7 m
Depth of the river, d2 = 0.85 m
Speed of the river, v1 = 1.2 m/s
Speed of the river, v2 = ?
Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²
Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²
To solve this question, we use a relation between the speed of the river and the volume of the river. We say,
Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have
32.4 * 1.2 = 4.93 * v2
38.88 = 4.93v2
v2 = 38.88 / 4.93
v2 = 7.89 m/s
Therefore, the speed of the river after the rapid is 7.89 m/s
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
