What is the heights part of Maine?

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago

You don't have to notify employees that a lockout/tagout is about to begin?

Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

3 0

2 years ago

A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength

Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0

2 years ago

Air at 25 C and 1 atm is flowing over a long flat plate with a velocity of 8 m/s. Determine the distance from the leading edge o

DIA [1.3K]

Answer:

Explanation:

Given data

Temprature of air= $25^{\circ}$

pressure =1 atm

velocity(V)=8m/s

From the table for air at $25^{\circ}$ and 1 atm

kinematic viscosity $\left ( \nu\right )=1.562\times 10^{-5} m^{2}/s$

Reynolds number for turbulent flow $=5\times 10^{5}$

and Re.no.= $\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X= $\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.976 m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.976}{\sqrt{5\times 10^5}}$

$\delta _x=6.901mm$

for water

kinematic viscosity $\left ( \nu\right )=8.91\times 10^{-7} m^{2}/s$

Reynolds number for turbulent flow $=5\times 10^{5}$

and Re.no.= $\frac{V \times X}{\nu }$

therefore length where turbulent flows start is

X= $\frac{\left ( Re.no.\right )\times \nu }{V}$

X=0.055m

Thickness of boundary layer is given by

$\delta _x=\frac{5X}{\sqrt{Re_x}}$

$\delta _x=\frac{5\times 0.055}{\sqrt{5\times 10^5}}$

$\delta _x=0.3889mm$

5 0

2 years ago

Read 2 more answers

A double-pipe heat exchanger is used to cool a hot fluid (cp = 3800 J/kg·K) entering the heat exchanger at 200°C with a flow rat

MAXImum [283]

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem

3 0

3 years ago