To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.
The volume of a tank is given by

Where
d = Diameter
h = Height
Considering that there are two stages, let's define the initial and final volume as,


We know as well by definition that

Then we have for the statement that


Replacing the previous data


Solving to get h,

Therefore the change is



Therefore te change in the height of the water in the tank is 0.37mm
Momentum = (mv).
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>
Answer:(10.69, 11.436)
Explanation:
Given
initial height of ball is 2 m
height of basket is 3.05 m
Launching angle

y=1.05
equation of trajectory of ball is given by

for x=12.27

u=10.69
for x=11.73

u=11.436 m/s
Thus range of speed is (10.69, 11.436)
Answer:
The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²
Explanation:
Data
Ta = 0.125 s
Tb = 0.08 s
Δtab = 0.5 s
distance = 1 cm
Process
1.- Calculate va
va = 1/0.125 = 8 cm/s
vb = 1/0.08 = 12.5 cm/s
2.- Calculate Δv
Δv = 12.5 - 8
Δv = 4.5 cm/s
3.- Calculate acceleration
a = Δv / Δt
a = 4.5/0.5
a = 9 cm/s²
Answer with Explanation:
We are given that
Weight of an ore sample=17.5 N
Tension in the cord=11.2 N
We have to find the total volume and the density of the sample.
We know that
Tension, T=
=buoyancy force
T=Tension force
W=Weight
By using the formula

N

Where
=Volume of object
=Density of water
=Acceleration due to gravity
Substitute the values then we get


Volume of sample=
Density of sample,
Where mass of ore sample=1.79 kg
Substitute the values then, we get

Density of the sample=