A block with mass m1m1m_1 is placed on an inclined plane with slope angle ααalpha and is connected to a second hanging block tha
1 answer:
Answer:
Explanation:
m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .
For equilibrium of m₁
m₁ g sinα= T + f where f is force of friction
m₁ g sinα= T + μsx m₁ g cosα
m₁ g sinα - μs x m₁ g cosα = T
For equilibrium of m₂
T = m₂g
Putting this value in equation above
m₁ g sinα - μs x m₁ g cosα = m₂g
m₂ = m₁ sinα - μs x m₁ cosα
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Answer:
.7934
Explanation:
Acceleration = change in velocity / change in time
A = 10.98 / 13.84
A = .7934
Answer:
1600 kJ/h per K, 888.88 kJ/h per °F and 888.88kJ/h per R
Explanation:
We make use of relations between temperature scales with respect to degrees celsius:
This means that a change in one degree celsius is equivalent to a change of one kelvin, while for a degree farenheit and rankine this is equivalent to a change of 1.8 on both scales.
So:
Answer:
v=1.295
Explanation:
What we are given:
a=5÷(3s^(1/3)+s^(5/2)) m/s^2
Start by using equation a ds = v dv
This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:
<em>a=2</em>
<em>b=1</em>
<em>x=5÷(3s^(1/3)+s^(5/2)) </em><em>m/s^2</em>
<em>dx=dv</em>
Integrate the left side the standard method.
<em>a=v</em>
<em>b=0</em>
<em>dx=dv</em>
<em>Integrating</em>
=v^2/2
Use Simpson's rule for the right site.
<em>a=b</em>
<em>b=a</em>
<em>x=f(x)</em>
f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)
If properly applied. you should now have the following equation:
v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]
=0.8376
Solve for v.
v=1.295