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
Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor, 
The static and dynamic friction coefficient is, 
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,
So, the acceleration of the car is 
. Hence, this is the required solution.
Answer:
physical health
Explanation:
Physical health has many components including: exercise, nutrition, sleep, alcohol & drugs, and weight management.
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
The wavelength of the 2nd harmonic is:
The wavelength of the 4th harmonic is:
It is not possible to find any integer n such that 
, therefore the correct options are A, B and D.
Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N
