To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
From the cinematic equations of motion we know that
Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
From the equation of motion where acceleration is equal to the velocity in function of time we have
Therefore the time required is 0.0705s
Answer:
Henri’s wave and Geri’s wave have the same amplitude and the same energy
Explanation:
The amplitude of a wave is the distance between the midpoint and the trough (or the crest). This is equivalent to half the distance between the trough and the crest. Therefore:
- amplitude of Henri's wave: 4 cm
- amplitude of Geri's wave: 8/2 = 4 cm
The energy of a wave is directly proportional to its amplitude.
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:
Kinetic energy, E = 133.38 Joules
Explanation:
It is given that,
Mass of the model airplane, m = 3 kg
Velocity component, v₁ = 5 m/s (due east)
Velocity component, v₂ = 8 m/s (due north)
Let v is the resultant of velocity. It is given by :
Let E is the kinetic energy of the plane. It is given by :
E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm
