Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7
<h3>How to solve for the value of s2 - s1</h3>
We have
= 


v2 = 5x2 - 3x2 + c
= 10-6+c
= 4+c

S2 - S1

= 6 + 6+c - 2+3+c
12+c-5+c = 0
7 = c
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Answer:

Explanation:
First we calculate the mass of the aire inside the rigid tank in the initial and end moments.
(i could be 1 for initial and 2 for the end)
State1


State2


So, the total mass of the aire entered is

At this point we need to obtain the properties through the tables, so
For Specific Internal energy,

For Specific enthalpy

For the second state the Specific internal Energy (6bar, 350K)

At the end we make a Energy balance, so

No work done there is here, so clearing the equation for Q



The sign indicates that the tank transferred heat<em> to</em> the surroundings.
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 ×
= 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 ×
/ 3
so d = 0.0133 m
so diameter is 14 mm
Answer:
a)We know that acceleration a=dv/dt
So dv/dt=kt^2
dv=kt^2dt
Integrating we get
v(t)=kt^3/3+C
Puttin t=0
-8=C
Putting t=2
8=8k/3-8
k=48/8
k=6
Answer:
Answer D
Explanation:
They need to be checked monthly