A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
1 answer:
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Answer:
6.3 m/s
Explanation:
m = mass of the block = 1.10 kg
k = spring constant of the spring
x = stretch in the spring = 0.2 m
t = time taken by block to come to zero speed first time = 0.100 s
T = Time period of oscillation
Time period of oscillation is given as
T = 2t
T = 2 (0.1)
T = 0.2 s
Time period is also given as
k = 1084.6 N/m
v = maximum speed of the block
using conservation of energy
Maximum kinetic energy = Maximum spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(1.10) v² = (1084.6) (0.2)²
v = 6.3 m/s
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:
Bulb 1 has more resistance.
Explanation:
Given that,
Two lightbulbs work on a 120-V circuit.
The power of circuit 1, P₁ = 50 W
The power of circuit 2, P₂ = 100 W
We need to find the bulb that has a higher resistance.
The power of the bulb is given by :
For bulb 1,
For bulb 2,
So, bulb 1 has higher resistance.