Answer: V = 675000v
Explanation:
Electric potential V can be calculated by using the formula
V = Kq/r
Where
V = electric potential
q = charge = 15µC
K = Coulomb constant = 9 × 10^9Nm^2C^-2
r = 20 cm
Substitute all the parameters into the formula
V = (9×10^9 × 15×10^-6)/20×10^-2
V = 135000/0.2
V = 675000 volt
Answer:
Final Length = 30 cm
Explanation:
The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:
F = kΔx
where,
F = Force applied
k = spring constant
Δx = change in length of spring
First, we find the spring constant of the spring. For this purpose, we have the following data:
F = 50 N
Δx = change in length = 25 cm - 20 cm = 5 cm = 0.05 m
Therefore,
50 N = k(0.05 m)
k = 50 N/0.05 m
k = 1000 N/m
Now, we find the change in its length for F = 100 N:
100 N = (1000 N/m)Δx
Δx = (100 N)/(1000 N/m)
Δx = 0.1 m = 10 cm
but,
Δx = Final Length - Initial Length
10 cm = Final Length - 20 cm
Final Length = 10 cm + 20 cm
<u>Final Length = 30 cm</u>
I think the answer would be D
but not sure, dont count on mre
Explanation:
Igneous rocks are formed by melting and cooling of magma originated from volcanic process.
when molten rock (rock liquefied by intense heat and pressure) cools to a solid state. Lava is molten rock flowing out of fissures or vents at volcanic centres (when cooled they form rocks such as basalt, rhyolite, or obsidian)
These rocks are strong, crystalline and dark in colour.
Answer:
a) Q = 80,000 cal
b) Q = 100,000 cal
c) Q = 540,000 cal
d) Q = 720,000 cal
Explanation:
a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion
= 1 * 80
= 80 kCal = 80,000 cal
b) 1 kg of O°C ice water to 1 kg of 100°C boiling water
Specific heat capacity, c = 1000cal/kg.C
![Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal](https://tex.z-dn.net/?f=Q_%7Bc%7D%20%3D%20mc%20%5Cdelta%20T%5C%5CQ_%7Bc%7D%20%3D%201%20%2A%201000%20%2A%20%28100%20-%200%29%5C%5CQ_%7Bc%7D%20%3D100000%20cal)
c) 1 kg of 100°C boiling water to 1 kg of 100°C steam
Latent heat of vaporization is needed for this conversion
![Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal](https://tex.z-dn.net/?f=Q_%7Bv%7D%20%3D%20ML_%7Bv%7D%20%5C%5CL_%7Bv%7D%20%3D%20540%20kCal%2Fkg%5C%5CQ_%7Bv%7D%20%3D1%2A%20540%20%5C%5CQ_%7Bv%7D%20%3D%20540%20kCal%20%3D%20540000%20cal)
d) 1 kg of O°C ice to 1 kg of 100°C steam.
Q = ![Q_{L} + Q_{c} + Q_{v}](https://tex.z-dn.net/?f=Q_%7BL%7D%20%2B%20Q_%7Bc%7D%20%2B%20Q_%7Bv%7D)
Q = 80,000 + 100,000 + 540,000
Q = 720,000 cal