We can base this on the equation of thermal expansion.
ΔL = L₀αΔT
where
ΔL is the expansion of length upon heating
L₀ is the initial length
α is the coefficient of linear expansion
ΔT is the temperature difference
So, if α for Aluminum is greater than α for Copper, then after heating, aluminum would be longer than copper.
Answer:
2 kg
Explanation:
Note: For the meter stick to be balanced,
Sum of clock wise moment must be equal to sum of anti clock wise moment
Wd = W'd' ................ Equation 1
Where W = weight of the rock, d = distance of the meter stick from the point of support, W' = weight of the that must be suspended for the meter stick to be balanced, d' = distance of the mass to the point of support.
make W' the subject of the equation
W' = Wd/d'............... Equation 2
Taking our moment about the support,
Given: W = mg = 1 ×9.8 = 9.8 N, d = 50 cm, d' = (75-50) = 25 cm
Substitute into equation 2
W' = 9.8(50)/25
W' = 19.6 N.
But,
m = W'/g
m = 19.6/9.8
m = 2 kg.
Answer:
C
Explanation:
To melt the alcohol
Heat needed = M . L = 2 . 25 = 50 kcal
To warm up the alcohol
Heat needed = M . sp. ht. . ∆t = 2 . 0.6 . 100 = 120 kcal
Total heat needed = 170 kcal
Assuming that 0.6 kcal/ kg / ˚C is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)
It’s true the acceleration of falling objects on earth due to gravity is 98ms2
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = 
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
