Answer:
The extension of the wire is 0.362 mm.
Explanation:
Given;
mass of the object, m = 4.0 kg
length of the aluminum wire, L = 2.0 m
diameter of the wire, d = 2.0 mm
radius of the wire, r = d/2 = 1.0 mm = 0.001 m
The area of the wire is given by;
A = πr²
A = π(0.001)² = 3.142 x 10⁻⁶ m²
The downward force of the object on the wire is given by;
F = mg
F = 4 x 9.8 = 39.2 N
The Young's modulus of aluminum is given by;
Where;
Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²
Therefore, the extension of the wire is 0.362 mm.
- The wavelength of the red light in "nanometer" is 7×
- Wavelength is given as : 7×
meter
- 1 nanometer = (
meter)
- Let X= value of the wavelength in nanometer.
1 nanometer = meter
X nanometer = 7× meter
- <em>If we Cross multiply</em>
X nanometer = (
)
X= 7× nanometer
Therefore, the wavelength in "nanometer" is 7×
Learn more at :brainly.com/question/12924624?referrer=searchResults
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Answer:
QC = 122 KJ
QH = 2.64 x 122 = 322 KJ
Explanation:
TH = 500 Degree C = 500 + 273 = 773 K
TC = 20 degree C = 20 + 273 = 293 K
W cycle = 200 KJ
Use the formula for the work done in a cycle
Wcycle = QH - QC
200 = QH - QC ..... (1)
Usse
TH / TC = QH / QC
773 / 293 = QH / QC
QH / QC = 2.64
QH = 2.64 QC Put it in equation (1)
200 = 2.64 QC - QC
QC = 122 KJ
So, QH = 2.64 x 122 = 322 KJ
