Answer:
Private jet is 200 mph
Commercial jet is 225 mph
Explanation:
Let the speed of private jet be x mph therefore the speed of commercial jet will be 2x-175 mph
The distance is the same
Speed is distance per unit time hence distance is product of speed and time
Distance by private jet is 9x
Distance by commercial jet is 8(2x-175)
Since the distance is the same
9x=8(2x-175)
9x=16x-1400
7x=1400
X=200
Then 2x-175= 400-175=225
Speed of private jet is 200 mph while commercial jet is 225 mph
When object travels with uniform velocity, no force acts on it. hence , yes.
Answer:
Explanation:
The period of oscillation is given as
T=2π√m/k
Making k subject of the formula
Square both sides of the equation
T²=4π²(m/k)
Cross multiply
T²k=4π²m
Then, divide through by T²
k=4π²m/T²
Where
k is spring constant
m is the mass of the bob
And T is the period of the oscillation
m=140g=0.14kg
14 oscillations takes 14 seconds
Then the period is
T=time/oscillation
T=14/14
T=1sec
Then,
k=4π²m/T²
k=4π²×0.14/1²
k=1.76N/m
Then, the spring constant is 1.76N/m
Answer:
a)v = 476.28 m / s
, b) T = 6.69 10⁵ N
, c) λ = 0.486 m
, d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m
Answer:
Mass of methane takne = 1.5g
moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles
mass of water = 1000 g
Initial temperature of water = 25 C
final temperature = 37 C
specific heat of water = 4.184 J /g C
1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules
2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J
3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules
This heat is released by 0.094 moles of methane
So heat released by one mole of methane =
- 622851.06 Joules = 622.85 kJ / mole
4) standard enthalpy of combustion = -882 kJ / mole
Error = (882-622.85) X 100 / 882 = 24.84 %