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STALIN [3.7K]
2 years ago
8

Engineers please help im not good when it comes to drawing​

Engineering
1 answer:
kakasveta [241]2 years ago
3 0

Answer:

I'm good at drawing and computer-animated design

Explanation:

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Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 90-gallon hot
Rudik [331]

Answer:

Q=36444.11 Btu

Explanation:

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Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

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Mass = Volume x density

1\ gallon = 0.13ft^3

Mass ,m= 90 x 0.13 x 62.36 lbm

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Q= 729.62 x 0.999 x (110-60) Btu

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5 0
3 years ago
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
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