If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
F=52*9.8=509.6
therefore the weight of the object is 509.6N
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
a) F = 2250 Ib
b) F = 550 Ib
c) new max force ( F newmax ) = 2850 Ib
Explanation:
A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up
max capacity of elevator = 24000 Ibs
counterweight = 1000 Ibs
To calculate the force (F) :
we first calculate the Tension using this relationship
Counterweight (1000) - T = ( 1000 / g ) ( g/4 )
Hence T = 750 Ib
next determine F
750 + F - 2400 = 2400 / 4
hence F = 2250 Ib
B ) calculate Tension first
T - 1000 = ( 1000/g ) ( g/4)
T = 1250 Ib
F = 2400 -1250 - 2400/ 4
F = 550 Ib
C ) determine design limit
Max = 2400 * 1.2 = 2880 Ib
750 + new force - 2880 = 2880 / 4
new max force ( F newmax ) = 2850 Ib
