Answer:
The density of the woman is 950.8 kg/m³
Explanation:
Given;
fraction of the woman's volume above the surface = 4.92%
then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%
the specific gravity of the woman
The density of the woman is calculate as;
Density of fresh water = 1000 kg/m³
Density of the woman = 0.9508 x 1000 kg/m³
Density of the woman = 950.8 kg/m³
Therefore, the density of the woman is 950.8 kg/m³
Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.
where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
Answer:
The tension in the rope is 262.88 N
Explanation:
Given:
Weight N
Length of rope m
Initial speed of ball
For finding the tension in the rope,
First find the mass of rod,
(
)
kg
Tension in the rope is,
N
Therefore, the tension in the rope is 262.88 N
Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above
b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr
The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz