A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
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Answer:
Explanation:
1. when both sides of the reactants and the products are equal
like h+oh ⇒ h20
hydrogen has 2 atoms on both sides and oxygen has one on both sides
2. no they are put to balance the equation
3. nope they are treated equally as all the other states
4. no if u are talking about a formula unit like NaCl for example it is aqueous not each element taken on its own if u are talking about just elements then i said before all states are treated equally
5. yup
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
Here, Fe = Fg
q.E = m.g
We have: E = 360 V
m = 5.4 × 10⁻⁵
g = 9.8 m/s² [ constant value for earth system ]
Substitute their values into the expression:
q (360) = 5.4 × 10⁻⁵ × 9.8
q = 52.92 × 10⁻⁵ / 360
q = -1.47 × 10⁻⁶ [ negative sign represents the nature of charge ]
So, Your Final answer would be 1.47 × 10⁻⁶
Hope this helps!
The velocity of the boy when he hits the water at the bottom of the slide is 14 m/s.
<h3>
Velocity of the boy at the bottom of the slide</h3>
The velocity of the boy when he hits the water at the bottom of the slide is calculated from the principle of conservation of energy.
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- h is height of the boy
- g is acceleration due to gravity
v = √(2 x 9.8 x 10)
v = 14 m/s.
Thus, the velocity of the boy when he hits the water at the bottom of the slide is 14 m/s.
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