For this assignment, you will construct a personal dietary plan that is reflective of your current diet in terms of its nutritional value.

Answers

Answer 1
Answer: you don’t do it and cheat

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What happened to a apple's weigh as the plane rose up toward the sky?

Answers


Okay, I don't know if this question is supposed to be a trick question or not. The weight of the apple does not change as the plane travels up the atmosphere, but the MASS changes. Weight doesn't change no matter what environment you're in, but the mass changes in different environments. In this case, the weight is constant but the mass is decreasing as you go higher up.

The copper wire and bulb is connected in a series with 220 V electric supply. Why only an electric bulb glows where as the copper wire remains the same.Give REASON to support the answer.

Answers

-- In a series circuit, the current ( I ) is the same at every point.

-- The power dissipated by any section of the circuit is I² x Resistance.

-- The wire has very low resistance, so I²R is very low dissipated power.

-- The filament in the bulb has most all of the resistance in the circuit,
so it dissipates virtually all the power of the circuit, and certainly much
more than the wires do.

An object is in uniform circular motion, tracing an angel at 30 degrees every 0.010 seconds. What's the period of this motion and how do I work it out?

Answers

Here's the rule you need to know
in order to answer this question:

                     1 full circle ==> 360 degrees .

Got that ?

Now you could set up a proportion:

     (30 degrees) / (0.01 second)  =  (360 degrees) / (time for full period)

Cross-multiply the proportion:

     (30°) · (period)  =  (360°) · (0.01 sec)

Divide each side by (30°) :    Period = (360° · 0.01 sec) / (30°)

                                                     =  (3.6° · sec) / (30°)

                                                     =  (3.6 / 30)  sec

                                                     =      0.12  sec .
___________________________________

Another way to look at it:

30°        takes    0.01 second
60°        takes    0.02 second
90°        takes    0.03 second
120°      takes    0.04 second
150°      takes    0.05 second
180°      takes    0.06 second
210°      takes    0.07 second
240°      takes    0.08 second
270°      takes    0.09 second
300°      takes    0.10 second
330°      takes    0.11 second
360°      takes   0.12 second

A train traveling 80.0 kph is blowing its horn as it approaches a railroad crossing. The horn has a frequency of 300.0 Hz. Assume the speed of sound is 331.5 m/s. What is the observed frequency of the horn?395 Hz

281 Hz

322 Hz

Answers

Answer:

322 Hz

Explanation:

v = speed of train approaching the railroad crossing = 80 km/h = 80 x 1000/3600 m/s = 22.22 m/s

V = speed of sound of the horn of train = 331.5 m/s

f = actual frequency of the sound from the horn = 300.0 Hz

f' = observed frequency of the horn

Using Doppler's effect, observed frequency is given as

f' = V f/(V - v)

inserting the values

f' = (331.5) (300.0)/(331.5 - 22.22)

f' = 322 Hz

the answer to your question is 322 hz
h

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Answers

Equation of motion:

y_(f)=y_(o)+v_(o)t+ (1)/(2) at^(2)

Since initial velocity is zero, the second term goes away:

y_(f)=y_(o)+0+ (1)/(2) at^(2)

y_(f)=y_(o)+(1)/(2) at^(2)

y_(f)-y_(o)= (1)/(2) at^(2)

y_(f)-y_(o)=5.4m

5.4m= (1)/(2) at^(2)

(2(5.4m) )/(a) = t^(2)

a = g = 9.81 (m)/( s^(2))

(2(5.4m) )/(9.81 (m)/( s^(2) ) ) = t^(2)

1.1 s^(2) = t^(2)\sqrt{1.1 s^(2)} = \sqrt{t^(2)}

t = 1.05s

Answer:

1.05 s

Explanation:

a crowbar having the lenght of 1.75 is used to balance a load of 500N if the distance between the fulcrum and the load is 0.5m find the effort applied to balance the load​

Answers

The effort applied = 200 N

Further explanation

Equilibrium :

\tt F_1.d_1=F_2.d_2

Total length = 1.75

The distance between the fulcrum and the load is 0.5 m ⇒ d₁=0.5 m

The distance between the fulcrum and the force applied :

\tt 1.75-0.5=1.25~m\rightarrow d_2=1.25~m

A load of 500N ⇒ F₁=500 N

The force applied :

\tt 500* 0.5=F_2* 1.25\n\nF_2=(500* 0.5)/(1.25)=200~N