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2 years ago

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How do I find applied force of an object given magnitude

1 answer:

Arte-miy333 [17]2 years ago

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What is electromagnetism

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The interaction of electric currents or fields and magnetic fields.

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Explain how the loss of a plant, or producer, in an ecosystem can affect a secondary consumer

vova2212 [387]

When plants die they become food for the decomposer for example Lettuce, If all the Lettuce in someones field died the decomposer would get its nutriets and make the soil stronger, so if you plant more Lettuce and it survives the Consumer would get to eat the lettuce. Hope this helped! :)

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2 years ago

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Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit

riadik2000 [5.3K]

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

$h_1 = 138.99 Btu/lbm$

steady flow energy equation is givena s

$h_1 + \frac{v_1^2}{2} = h_2 + \frac{v_2^2}{2}$

$138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}$

$h_2 = 94.344 Btu/lbm$

At 90 lb/in2 Tsat = 72.78 degree F

$h_f = 35.715 Btu/lbm$

hfg = 77.345 Btu/lbm

h = hf + x hfg

$94.344 = 35.715+ x \times 77.345$

solving for x we get

x = 0.75801 = 75.801%

$T_2 = 72..78 degree F$

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2 years ago

Because not all stars are the same distance from earth, it can be difficult to determine?

Inessa [10]

I think answer is A. The Brightness of a Star

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A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient

Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

conservation of energy, for this case manifested through work and kinetic energy.

$W = \Delta KE$

$W = F*d$

Where,

F= Force (Frictional at this case $F_r = \mu N$ )

d= Distance

$\Delta KE = \frac{1}{2} mv^2$

Where,

m = mass

v = velocity

Equation both terms,

$F*d = \frac{1}{2}mv^2$

$\mu mg *d = \frac{1}{2}mv^2$

$\mu g * d = \frac{1}{2}v^2$

$d = \frac{1}{2} \frac{v^2}{\mu g}$

Replacing with our values we have that

$d = \frac{1}{2} \frac{25^2}{0.65*9.8}$

$d = 49.05m$

Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m

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2 years ago