Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
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Answer:
a) P =392.4[Pa]; b) F = 706.32[N]
Explanation:
With the input data of the problem we can calculate the area of the tank base
L = length = 10[m]
W = width = 18[cm] = 0.18[m]
A = W * L = 0.18*10
A = 1.8[m^2]
a)
Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:
P = density * g *h
where:
density = 1000[kg/m^3]
g = gravity = 9.81[m/s^2]
h = heigth = 4[cm] = 0.04[m]
P = 1000*9.81*0.04
P = 392.4[Pa]
The force can be easily calculated knowing the relationship between pressure and force:
P = F/A
F = P*A
F = 392.4*1.8
F = 706.32[N]
Answer:
The third drop is 0.26m
Explanation:
The drop 1 impacts at time T is given by:
T=sqrt(2h/g)
T= sqrt[(2×2.4)/9.8]
T= sqrt(4.8/9.8)
T= sqrt(0.4898)
T= 0.70seconds
4th drops starts at dT=0.70/3= 0.23seconds
The interval between the drops is 0.23seconds
Third drop will fall at t= 0.23
h=1/2gt^2
h= 1/2×9.81×(0.23)^2
h= 0.26m
Answer:
The compression in the spring is 5.88 meters.
Explanation:
Given that,
Mass of the car, m = 39000 kg
Height of the car, h = 19 m
Spring constant of the spring, 
We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,
x is the compression in spring
So, the compression in the spring is 5.88 meters.
Answer:
The angular acceleration of the pencil<em> α = 17 rad·s⁻²</em>
Explanation:
Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:
τ = I α (1)
W r = I α (2)
The weight is that the pencil has is,
sin 10 = r / (L/2)
r = L/2(sin(10))
The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:
I = 1/3 M L²
Thus,
mg(L / 2)sin(10) = (1/3 m L²)(α)
α(f) = 3/2(g) / Lsin(10)
α = 3/2(9.8) / 0.150sin(10)
<em> α = 17 rad·s⁻²</em>
Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>
