W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
Im not for sure but i think it takes a couple hundred years (or according to the climate)
The correct answer to the question is vertically downward i.e towards the centre of earth.
EXPLANATION:
As per the question, the box is pulled to the right.
Hence, the direction of the applied force is towards right.
We are asked to determine the direction of the gravitational force that acts on the body.
Before answering this question, first we gave to understand the gravitational force of earth.
Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre. It is equivalent to the weight of the body.
The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.
Hence, the direction of the gravitational force which acts on the box is vertically downward.
To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:
Where,
P = Axial Load
l = Gage length
A = Cross-sectional Area
E = Modulus of Elasticity
Our values are given as,
l = 3.5m
D = 0.028m
E = 200GPa
Replacing we have,
Therefore the change in length is 1.93mm
Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736
Explanation:
According to the Arrhenius equation,
or,
![\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 525K
= rate constant at 545K
= activation energy for the reaction = 185kJ/mol= 185000J/mol (1kJ=1000J)
R = gas constant = 8.314 J/mole.K
= initial temperature = 525 K
= final temperature = 545 K
Now put all the given values in this formula, we get
![\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%29%3D%5Cfrac%7B185000J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B525K%7D-%5Cfrac%7B1%7D%7B545K%7D%5D)
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736
