Answer:
a) 11.2 g
b) 3.73 g.
Explanation:
a) If we assume temperature of mixture to be 100°C , heat released by steam will be 11.2 x 540 = 6048 cals and heat gain gained by will be
79 x 80 + 79 x 1 x 100 = 14220 cals . Since former heat is less than later heat ,water will not be warmed up to 100°C. Let equilibrium temperature be t .
Heat gained by water = 79 x 80 + 79 x 1 x t = 11.2 x 540 + 11.2( 100 - t )
t = 9.4°
amount of steam condensed = 11.2 g.
b) In this case, whole of water will be warmed up to 100°C as steam is much .heat required by water to warm up to boiling point
= 11.2 x 80 + 11.2 x 100 = 2016 cals
amount of steam condensed = 2016 / 540 = 3.73 g .
A change in velocity or a change in direction, or both! Hope this helps
Explanation:
In induction, what charge does a neutral substance gain compared to the object brought near it?
The neutral object gains the same type of charge as the object that touched it because the electrons move from one object to the other (Figure 10.16). Induction is the movement of electrons within a substance caused by a nearby charged object, without direct contact between the substance and the object.
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
Complete Question
An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r
Answer:
The difference is 
Explanation:
From the question we are told that
The radius of the soap bubble is 
The potential of the soap bubble is 
The new radius of the soap bubble is 
The initial electric potential is mathematically represented as
The final electric potential is mathematically represented as
The initial potential is mathematically represented as
The final potential is mathematically represented as
Now
substituting values
=> 
So
Therefore
![U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]](https://tex.z-dn.net/?f=U_f%20-U_i%20%3D%20%20%20%20%20%5Cfrac%7BV_1%5E2%7D%7B2k%7D%20%5B%5Cfrac%7B%20R_1%20%7D%7B%20%2A%200.19%5E2%7D%20-%20R_o%5D)
where k is the coulomb's constant with value 
substituting values
![U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]](https://tex.z-dn.net/?f=U_f%20-U_i%20%3D%20%20%20%20%20%5Cfrac%7B307%5E2%7D%7B9%20%2A%2010%5E%7B9%7D%7D%20%5B%5Cfrac%7B%200.014155%20%7D%7B%200.19%5E2%7D%20-%200.0745%5D)
