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Varvara68 [4.7K]
4 years ago
15

You build a 2.0-kg mound of mashed potatoes on a kitchen scale (which is a spring scale). You then use a large spoon to press on

the top of the mound at a constant force of 3.0 N. While you are doing this, the scale reading increases by 5.0%. (a) What is the acceleration of the center of mass of the potatoes? (b) Describe what happens to the shape of the mound and the scale reading as time goes by.

Physics
2 answers:
igor_vitrenko [27]4 years ago
8 0

Answer:

a) a=1m/s^2 downwards

b) The mound will be reduced in height (the acceleration is downwards) and will begin to grow sideways. The reading will remain the same until the spoon reaches the scale surface.

Explanation:

On the mound, the forces acting before the spoon are:

N1 - m*g = 0   Where N1 is equivalent to the reading of the scale.

N1 = 2*10 = 20N

After the spoon starts pressing, this value increases by 5%:

N2 = 20*105% = 21N  So, at this point the forces on the moud are:

N2 - m*g - F = m*a   Solving for a:

a = -1m/s^2   This result tells us that the center of mass of the potatoes is moving downwards. Since it cannot move because of the scale, it has no other option than to grow sideways as it is pushed.

marshall27 [118]4 years ago
5 0

Answer:

a= -1.01m/s^2

Explanation:

A) We define all the variables that are given,

In this way

Acceleration = a

m = 2.0Kg

F = 3.0N

We obtain the Normal assuming 5% of the potato reading scale, so

N = (2 + 2 * 5\%) * 9.8

N = 20.58N

We add forces to reach equilibrium, and you get

N-F-mg = ma

** The notation of the symbols is done according to the notation of the graph

So,

a = \frac {N-F-mg} {m}

a = \frac {20.58-3-2 * 9.8} {2}

a = -1.01m / s ^ 2

B) When making pressure it is understood that the shape takes the perpendicular direction of the direction of the force that is performed. If the pressure remains constant, the scale will also increase in its measurement.

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an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

E1= k*q1/(h1-h2)..............(1)

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1


[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2\\

E1=[tex]E= 359502+359502\\E=719004 V/mV/m[/tex]

similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0
3 years ago
A jet - powered car called the spirit of America required 9600meters to stop from its highest speed . If the car decelerated at
Mazyrski [523]
v^{2} =  u^{2}  +  2ar
0 = u^2 + 2*(-2)*9600
u^2 = 38400
u = 195.96 m/s
6 0
3 years ago
A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. T
slega [8]

Answer:

 N_s\approx41667 \hspace{3}lo ops

Explanation:

In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:

\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings

In this case:

V_p=120V\\V_s=100kV=100000V\\N_p=50

Therefore, using the previous equation and the data provided, let's solve for N_s :

N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps

Hence, the number of loops in the secondary is approximately 41667.

3 0
3 years ago
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride
marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

\Delta x = 6.17 cm = 0.0617 m

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

2mg = k'\Delta x (1)

where

m = 76.2 kg is the mass of each children

g=9.8 m/s^2 is the acceleration of gravity

k' is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

k'=k+k=2k

Substituting into (1) and solving for k, we find:

2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m

b)

The period of the oscillating system is given by

T=2\pi \sqrt{\frac{m}{k'}}

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

k'=2k=2(12,103)=24,206 N/m is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

f=\frac{1}{T}

where

f is the frequency

T is the period

In  this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

f=\frac{1}{2.09}=0.478 Hz

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

t=10T=10(2.09)=20.9 s

#LearnwithBrainly

7 0
3 years ago
Hi
Vera_Pavlovna [14]

1) the cotton bolls burst open, and the white cotton fiber can be seen. Ginning: The cotton picked up from the plants has seeds in it. The process of removing cotton seeds from pods is called ginning. Ginning was traditionally done by hand.

2) Preparation

To be used for thread or fabric, raw seed cotton must cleaned and free of debris. Seeds, burrs, dirt, stems and leaf material are removed from the cotton during ginning.

Module feeders break the modules apart and feed the cotton into the gin.

Some module feeders have giant conveyors moving the modules into stationary heads.

Other module feeders are self-propelled, moving down a track alongside the modules.

Once in the cotton gin, the seed cotton moves through dryers and through multiple cleaning machines that remove the gin waste such as burrs, dirt, stems and leaf material from the cotton. Once thoroughly clean and dry, the cotton goes to the gin stand where circular saws with small, sharp teeth pluck the fiber from the seed. This process makes the sliver smoother so more uniform yarns can be produced. Then the cotton is packed tightly into bales, ready to be processed into textiles.

Even though the cotton is cleaned during the ginning process, it's not nearly as clean as  it needs to be. Cotton fibers are shaved from the bales and sent through a series of cleaning and drying machines. The mixed and fluffed-up cotton goes into a carding machine which finishes the cleaning and straightening of the fibers, making them into a soft, untwisted rope called a sliver (pronounced sly-ver).

The sliver is drawn out to a thinner strand and given a slight twist to improve strength, then wound on bobbins (spools wound with the thread-like product for storage).  

It is now called roving and the roving bobbins are now ready for the spinning process.

Spinning

On modern spinning frames, yarn is mare directly from the sliver. The spinning devices take fibers from the sliver and rotate it up to 2,500 revolutions in a second twist that makes fibers into a yarn for weaving or knitting into fabrics.

Spinning is the last process in yarn manufacturing. Today's mills draw and twist the roving into yarn and place it on bobbins. They do this quite efficiently. A large, modern mill can produce enough yarn or thread in 30 days to wrap around the earth 2300 times or go to and return from the moon 235 times. With the use of automatic winding, the yarn bobbins are transferred to larger bobbins called cheese cones. These cheese cones can be stored until they are needed in the weaving process.

Weaving

Looms weave cotton yarns into fabrics the same way the first hand-weaving frames did. Modern looms work at great speeds, interlacing the length-wise yarns (warp) and the crosswise yarns (weft).

Warp refers to yarns that run lengthwise in woven goods. In preparation of warp yarns for weaving, hundreds of yarn strands are wound from cheese cones onto a large warp beam. Yarns on this beam are then coated with a starch mixture (sizing compound) to add strength for weaving.

Weft is the yarn that runs crosswise in woven goods and may be referred to as filling yarn. Sizing is not placed on weft because flexibility is needed in the weaving process.

In today's most modern mills, the weft is fed into the loom from cheese cones with air-jets at such a high speed that its movement cannot be seen.

The woven fabric, called gray goods, is sent to a finishing plant where it is bleached, pre-shrunk, dyed, printed and given a special finish before being made into clothing or products for the home. Other machines make knits for use in shirts, sweaters or blankets.

this website will help you: https://www.quilting-in-america.com/process-of-making-cotton.html

i hope this helps :)

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3 years ago
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