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Arturiano [62]
3 years ago
7

At the instant that the velocity of the crate is v⃗ =(3.40m/s)ι^+(2.20m/s)j^, what is the instantaneous power supplied by this f

orce?
Physics
1 answer:
Zigmanuir [339]3 years ago
5 0
I found some missing information about this problem online. We are given the force:
F = F =(-7.50N)i +(3.00N)j
Power is defined as the rate of doing work. 
This is the formula:
P= \frac{dW}{dt}
Where P is power, W is work. 
Work is defined as:
W=F\cdot r
F is the force and r is the displacement.
If we assume that force is not changing (it's constant) with time we get:
P= \frac{dW}{dt}=F \frac{dr}{dt}=F\cdot v
Keep in mind that both force and velocity are vectors, so we have to multiply each component separately.
Finally, we get:
P=F_i\cdot v_i+F_j\cdot v_j=(-7.50N)(3.40\frac{m}{s})+(3.00N)(2.20\frac{m}{s})=
-18.9 W


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Read 2 more answers
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

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3 years ago
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