Answer:
The Answer is gonna be B.
Air warms and rises.
To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as
Where,
T = Temperature
R = Gas ideal constant
n = Number of moles in grams.
Our values are given as
The temperature is
Therefore the root mean square velocity would be
The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity
Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736
Answer:i=300 mA
Explanation:
Given
inductance(L)=40 mH
Resistor(R)=
Voltage(V)=15 V
Time constant()=
current
Current as a function of time is given by
i= 299.95 mA
To replace all forces on a pipe by the equivalent force
T= 8× 50 = 400lb.in
M = 16.30 = 480lb.in
F₂ = 50lb
To calculate the polar moment of inertia of shaft is
J = π/π²×(R⁴-r⁴)
= π²/2×(0.875⁴ - 0.750⁴)
= 0.423in.³
To calculate the moment of inertia
J= 1/2(J)
=1/2 (0.423)
=0.21188in.³
To calculate shear flow
Qy= 2/3(R³-r³)
= 2/3(0.875³- 0.750³)
= 0.16536in.³
To calculate the thickness of the shaft
t= R-r = 0.875 - 0.750 = 0.125 in.
The stress due to torsions is.
Tx = TR/J = 400 × 0.875/0.42376
=825. 9psi.
The stress due to bending
Qx =My/T = 480 × 0.875/0.2118
=1982.3psi
The stress due to transverse shear
Qx = VQ/I(2t)
=50 × 0.16536/0.2118× 0.250
=156.1psi
Answer:
The initial velocity of the bullet is 1066.63 m/s
Explanation:
The problem involves two different events: A collision between two objects and the stretching of a spring
Since we know the final condition of the system, we start from the end and back to the initial condition
The system of the bullet+block with mass= 4Kg+8gr = 4.008 Kg stretches the spring a distance of x=0.087 m
The work done when stretching a spring of constant k is
This work is equivalent to the change of potential energy of the spring, which is the result of the transformation of the kinetic energy of the total mass of the system
Solving for v' (the velocity of the system after the collision took place)
This is the resulting velocity after the bullet lodged into the block. That event complies with the law of conservation of linear momentum
Since the block is initially at rest ()
Solving for :
The initial velocity of the bullet is 1066.63 m/s