Answer:
0.44
Explanation:
Poisson's ratio is:
ν = (3K − E) / 6K
where K is the bulk modulus and E is Young's modulus.
Young's modulus is:
E = FL / (AΔL)
where F is the force, L is the initial length, A is the cross sectional area, and ΔL is the change in length.
E = (20 kg × 9.8 m/s²) (1.7 m) / (π (0.02 m)² × 0.0005 m)
E = 0.530×10⁹ Pa
Bulk modulus is:
K = -ΔP / (ΔV/V)
where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.
K = -(180 atm × 101325 Pa/atm) / (-0.012)
K = 1.52×10⁹ Pa
Therefore, the Poisson's ratio is:
ν = (3(1.52×10⁹ Pa) − 0.530×10⁹ Pa) / 6(1.52×10⁹ Pa)
ν = (3(1.52) − 0.530) / 6(1.52)
ν = 0.442
Rounded to 2 significant figures, the Poisson's ratio is 0.44.
Answer:
a) 390J
b) 322J
c) 68J
Explanation:
We need to calculate the power given by the battery. the power is given by:
Watts is J/s so:
The thermal energy in the wire is given by:
And the the dissipated thermal energy in the battery will be the remainig energy:
Answer:
The frequency of a wave is measured in Hz (hertz)
Answer:
15.64 KN
Explanation:
mass of the elevator cab with a single occupant= 2300 kg
acceleration relative to the cab 
= 6.80 m/s^2
acceleration of the coin relative to the cab in the opposite direction of motion of cab so we can consider it as a= -6.80 m/s^2
The acceleration of elevator cab relative to the ground 
now we can say that
=-6.80+ a_{eg}= -9.8
[tex]a_{eg}=-9.8+6.80=-3.8
The forces that act on elevator cab are tension and gravitational, applying newtons second law
T- mg= ma_{eg}
Then the tension in the cable is
T= 2300(-3.8)+2300×9.8= 15640 N= 15.64 KN
therefore tension in the string will be 15.64 KN
Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
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We assume the launch direction is either up or down.
