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Yuri [45]
3 years ago
7

your friend want to know the amont of salt added to her favorite fast food french fries what would you recommend that she do to

find out the amount of salt
Physics
1 answer:
Alexeev081 [22]3 years ago
5 0
That nutritional information is usually not posted up on the wall,
but by federal law, it's available on request, to any customer, at
any fast-food place. 

All she has to do is ask to see the manager, and then ask the
manager for the nutritional information booklet. 

She'll get a nice printed booklet that tells the salt content in the
fries, AND the amount of many other substances in ALL the items
sold at that place.

It's the law.  Don't be bashful.
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To protect her new two-wheeler, Iroda Bike
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Answer:

The length of chain she is allowed is 1.169 ft

Explanation:

The given parameters are;

The linear density of the chain = 0.83 lb/ft

The weight limit of the chain she wants = 1.4 lb

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Therefore, in order to keep the chain below 1.4 lb, we have;

Linear density = Weight/length

Therefore;

The maximum length she wants = Weight/(Linear density)

Which gives;

The maximum length she wants = 1.4 lb/(0.83 lb/ft) =1.169 ft

Therefore;

The length of chain she is allowed = 1.169 ft.

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3 years ago
The difference between the frequency fff and the frequency ωωomega is that fff is measured in cycles per second or hertz (abbrev
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Answer:

Radians

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Please help meeeeeeeeeeeeeeee
maxonik [38]

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1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
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Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

   PE = (2.5)(98)(14) and

   PE = 340 J

b. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 and

   KE = 250 J

c. TE = KE + PE so

   TE = 340 + 250 and

   TE = 590 J

d. PE at 8.7 m:

   PE = (2.5)(9.8)(8.7) and

   PE = 210 J

e. The KE at the same height:

   TE = KE + PE and

   590 = KE + 210 so

   KE = 380 J

f. The velocity at that height:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } so

   v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

   590 = KE + PE and

   PE = (2.5)(9.8)(11.6) so

   PE = 280 then

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   v=\sqrt{\frac{2(310)}{2.5} } so

   v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

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   26 = 0 + 9.8t and

   26 = 9.8t so the time at 26 m/s is

   t = 2.7 seconds. Now we use that in the equation for displacement:

   Δx = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 so

   Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

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