Answer:
1.54 kg
Explanation:
mass of first block (m) = 0.76 kg
acceleration due to gravity (g) = 9.8 m/s
what is the mass (m) of the second block
mg = kx
where m is the mass, g is the acceleration due to gravity, k is the
spring constant and x is the extension
0.76 x 9.8 = kx
7.5 = kx
k = 7.5/x ... equation 1
- when a second block is attached to the first one the amount of stretch triples (this means that extension (x) = 3x)
therefore the new mass becomes m + 0.76 and the extension
becomes 3x
with the new mass and extension, mg = kx now becomes
(m+0.76)g = k(3x) ... equation 2
Recall that k = 7.5/x from equation 1, substituting this value of k into
equation 2 we have
(m+0.76)g =
× (3x)
(m+0.76)g = 7.5 × 3
substituting the value of g = 9.8 m/s^{2}
(m + 0.76) x 9.8 = 7.5 x 3
m + 0.76 = 22.5 ÷ 9.8
m + 0.76 = 2.3
m = 2.3 - 0.76 = 1.54 kg
Answer:
The voltage across a semiconductor bar is 0.068 V.
Explanation:
Given that,
Current = 0.17 A
Electron concentration 
Electron mobility 
Length = 0.1 mm
Area = 500 μm²
We need to calculate the resistivity
Using formula of resistivity


Put the value into the formula


We need to calculate the resistance
Using formula of resistance



We need to calculate the voltage
Using formula of voltage

Put the value into the formula


Hence, The voltage across a semiconductor bar is 0.068 V.