Answer: ![-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BBr%5E.%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Explanation:
Rate of a reaction is defined as the rate of change of concentration per unit time.
Thus for reaction:
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{d[Br^.]}{2dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D)
or ![Rate=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Thus ![-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Answer:
Bulk modulus = 1.35 × 
Pa
Explanation:
given data
density = 1400 kg/m³
frequency = 370 Hz
wavelength = 8.40 m
solution
we get here bulk modulus of the liquid that is
we know Bulk Modulus = 
...............
here 
is density i.e 1400 kg/m³
and v is = frequency × wavelength
v = 370 × 8.40 = 3108 m/s
so here bulk modulus will be as
Bulk modulus = 3108² × 1400
Bulk modulus = 1.35 × Pa
6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
