Answer:
The range is maximum when the angle of projection is 45 degree.
Explanation:
The formula for the horizontal range of the projectile is given by
The range should be maximum if the value of Sin2θ is maximum.
The maximum value of Sin2θ is 1.
It means 2θ = 90
θ = 45
Thus, the range is maximum when the angle of projection is 45 degree.
If the angle of projection is 0 degree
R = 0
It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.
If the angle of projection is 30 degree.
R = 0.088u^2
If the angle of projection is 45 degree.
R = u^2 / g
Answer:
0.106
Explanation:
For 1 liter of diesel the car can get 19 km, if it takes 0.2 MJ for each km then it would take the total energy of 19*0.2 = 3.8 MJ to move an aerodynamic car 19 km. Since 1 liter of of diesel also contains 36 MJ in internal energy, then the efficiency of the diesel engine is the ratio of its output energy over its input energy:
Answer:
The equilibrium temperature is
21.97°c
Explanation:
This problem bothers on the heat capacity of materials
Given data
specific heat capacities
copper is Cc =390 J/kg⋅C∘,
aluminun Ca = 900 J/kg⋅C∘,
water Cw = 4186 J/kg⋅C∘.
Mass of substances
Copper Mc = 235g
Aluminum Ma = 135g
Water Mw = 825g
Temperatures
Copper θc = 255°c
Water and aluminum calorimeter θ1= 16°c
Equilibrium temperature θf =?
Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water
McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)
Substituting our data into the expression we have
235*390(255-θf)=
(135*900+825*4186)(θf-16)
91650(255-θf)=(3574950)(θf-16)
23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6
Collecting like terms and rearranging
23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf
8.2*10^6=3.66*10^6θf
θf=80.5*10^6/3.6*10^6
θf =21.97°c
Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
- 
L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
P = ρRT/M
Explanation:
Ideal gas equation is given as follows generally:
PV = nRT (1)
P = pressure in the containing vessel
V = volume of the containing vessel
n = number of moles
R = gas constant
T = temperature in K
n = m/M
m = mass of the gas contained in the vessel in g
M = molar mass in g/mol
ρ = m/V
Density of the gas = ρ
Substituting for n in (1)
PV = mRT/M. (2)
Dividing equation (2) through by V
P = m/V ×RT/M
P = ρRT/M
