Answer:
The movable piston
Explanation:
Work is said to be done when a distance is been covered by a force . In this case kinetic energy will be change by an equal amount into work done.
Pushing the piston with a known mass of (m) and an accelarating rate from rest of ( a) to cover a known distance of (d).The idea of work done is been achieved and can be mathematically represented by:
- Work done = Force x distance (d)
- Force = mass (m) x acceleration (a)
Answer:
35mA
Explanation:
Hello!
To solve this problem we must use the following steps
1. Find the electrical resistance of the metal rod using the following equation

WHERE
α=
metal rod resistivity=2x10^-4 Ωm
l=leght=2m
A= Cross-sectional area

solving

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)
3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy
I1+I2=5
4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)
V1=V2
(0.14I1)=2000(i2)
solving for i1
I1=14285.7i2
5.Now we use the equation found in step 3
14285.7i2+i2=5

Answer: Mass of the planet, M= 8.53 x 10^8kg
Explanation:
Given Radius = 2.0 x 106m
Period T = 7h 11m
Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.
This is represented by the equation
T^2 = ( 4π^2/GM) R^3
Where T is the period in seconds
T = (7h x 60m + 11m)(60 sec)
= 25860 sec
G represents the gravitational constant
= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet
Making M the subject of the formula,
M = (4π^2/G)*R^3/T^2
M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2
Therefore Mass of the planet, M= 8.53 x 10^8kg
Answer:
Volume of gasoline that expands and spills out is 1.33 ltr
Explanation:
As we know that when temperature of the liquid is increased then its volume will expand and it is given as

here we know that

volume expansion coefficient of the gasoline is given as

change in temperature is given as


Now we have


Answer:
Plot ln K vs 1/T
(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol
Explanation:
This is an example of the Arrhenius equation:

Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
Data:

Calculations:
(a) Rise
Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004
(b) Run
Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹
(c) Slope
Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹
(d) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol