On the whole, the metals burn in oxygen to form a simple metal oxide. Beryllium is reluctant to burn unless it is in the form of dust or powder. Beryllium has a very strong (but very thin) layer of beryllium oxide on its surface, and this prevents any new oxygen getting at the underlying beryllium to react with it.
Substance A would have a delta T (change in temp) rise 1/2 the rise in substance B.
Q=mc x delta T
Q= heat energy in Joules
m= mass of substance heated or cooled
c= specific heat
ΔT is change in temp.
Solve for change in temp=. Q/mc
Specific heat and mass are not inversely proportional to heat energy input.
Putting into real world scenario of using water to heat a building.
Specific heat of water is 1.
It takes 1 btu to raise one pound of water 1 degF. at a base of 60 degF
Acetone specific heat is .51
So it takes half the amount of heat input to get a 100 degree ΔT, as compared to water.
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
Answer:
Explanation:
a. The source of centripetal force on the car is (3) the static friction force.
b. ac = v²/R = (20²)/50 = 8 m/s²
c. Fc = m(ac) = 1500(8) = 12 kN
d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82
Answer:
F= 5.71 N
Explanation:
width of door= 0.91 m
door closer torque on door= 5.2 Nm
In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.
so wee need to exert 5.2 Nm torque on the door.
If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.
T= r x F
T= r F sin∅
F= T/ (r * sin∅)
F= 5.2/ (0.91 * 1)
F= 5.71 N
