A ball on a cart is moving at a rate of 2 m/s. The cart suddenly stops and the ball continues to travel in the same direction at
2 answers:
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This question is incomplete, the complete question is;
A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Answer:
the required heat energy is 16 J
Explanation:
Given the data in the question;
Lets consider the ideal gas equation;
PV = nRT
from the image, we calculate initial pressure;
Pi = ( 2000N/M × 0.06m) / 0.0008 m²
Pi = 15 × 10⁴ Pa
next we find Initial velocity
Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²
now we find the number of moles
n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K
N = 6.6 × 10⁻³ mol
next we calculate the final temperature;
Pf = ( 2000N/m × 0.08) / 0.0008 m²
Pf = 2 × 10⁵ Pa
Calculate the final Volume
Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³
we also determine the final temperature
= (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK
= 466.8 K
so change in temperature ΔT
ΔT = 466.8 K - 300K = 166.8 K
we then calculate the change in thermal energy
ΔU = nCΔT
ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K
ΔU = 13.761 J
C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic
now we calculate the work done;
W = 1/2 × K( x² - x
² )
W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )
= - 2.8 J
and we then calculate the heat energy using the following expression;
Q = ΔU - W
we substitute
Q = 13.761 - (- 2.8 J)
Q = 13.761 + 2.8 J)
Q = 16 J
Therefore, the required heat energy is 16 J
Answer:
23. 4375 m
Explanation:
There are two parts of the rocket's motion
1 ) accelerating (assume it goes upto h1 height )
using motion equations upwards
Lets find the velocity after 2.5 seconds (V1)
V = U +at
V1 = 0 +5*2.5 = 12.5 m/s
2) motion under gravity (assume it goes upto h2 height )
now there no acceleration from the rocket. it is now subjected to the gravity
using motion equations upwards (assuming g= 10m/s² downwards)
V²= U² +2as
0 = 12.5²+2*(-10)*h2
h2 = 7.8125 m
maximum height = h1 + h2
= 15.625 + 7.8125
= 23. 4375 m
Answer:
The compression in the spring is 5.88 meters.
Explanation:
Given that,
Mass of the car, m = 39000 kg
Height of the car, h = 19 m
Spring constant of the spring,
We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,
x is the compression in spring
So, the compression in the spring is 5.88 meters.