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3241004551 [841]
2 years ago
7

An elevator is used to either raise or lower sacks of potatoes. In the diagram, a sack of potatoes of mass 10 kg is resting on a

scale that is resting on the floor of an accelerating elevator. The scale reads 12 kg. What is the acceleration of the elevator?
Physics
1 answer:
vova2212 [387]2 years ago
8 0
Technically, if your question refers us to a diagram, then you owe us a peek at the
diagram. But there may be enough info in the description to solve this one blind.

Newton's second law of motion: F = M A
"The force on an object is the product of the object's mass and its acceleration."

Divide each side by 'A': A = F / M
"The rate at which an object accelerates is its mass divided by the force on it."
Hold that thought for a moment, while we go off on a tangent:
================================
Everybody talks about "kg" as if it were a force, and this question shows why
that's a terrible thing to do.  In this question, "kg" is used BOTH as a mass AND
as a force. I can't think of a better way to confuse students who are just now
working with this stuff for the first time. The question is badly written, although,
in the real world, scales do read in 'kg'.
"Kg" is a mass. It is not a force.

I think that rather than try to teach more physics to get out of this hole,
the best way to go at it is like this:

If the scale were sitting still, on the ground, or rising at a steady rate and
not accelerating, then the only force on the 10 kg mass would be the force
of gravity, and the scale would read '10 kg'. But in the upward-accelerating
elevator, the scale reads 20% more, telling us that 20% more than the force
of gravity is acting on the mass. That extra 20% of upward force is provided
by the upward acceleration, which must be 20% of the acceleration of gravity.
The acceleration of gravity is 9.8 meters per second² .
The elevator is accelerating (0.2 x 9.8) = <u>1.96 meters per second²</u>.
 


.

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We are given the following values:

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energy E = 3,000 J

 

The formula for potential energy is:

E = w h

where h is the height the person has to climb, therefore:

h = 3000 / 1067.52

<span>h = 2.81 m</span>

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<span>Therefore he has to climb 2.81 meters</span>

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A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n
Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

\lambda=\frac{v}{f}

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B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

f=\frac{v}{\lambda}

where for the dolphin:

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Substituting into the equation,

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C. 13.6 m

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\lambda=\frac{v}{f}

where

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f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

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\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz

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Polerization is the anwser

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Read 2 more answers
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

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(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

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Suppose Earth's mass increased but Earth's diame-
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Answer: It would increase.

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Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0
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