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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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Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m
OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

E_m = \frac{1}{2} m\omega_1^2A_1^2

The energy of the system having mass 2m is,

E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2

For the two expressions mentioned above remember that the variables mean

m = mass

\omega =Angular velocity

A = Amplitude

The energies of the two system are same then,

E_m = E_{2m}

\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2

\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}

Remember that

k = m\omega^2 \rightarrow \omega^2 = k/m

Replacing this value we have then

\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}

But the value of the mass was previously given, then

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}

\frac{A_1}{A_2} = 1

Therefore the ratio of the oscillation amplitudes it is the same.

5 0
2 years ago
A projectile is launched at an angle of 45° from the horizontal and lands 21 s later at the same height from which it was launch
irinina [24]

Answer:

a) initial speed of projectile = 145.5 m/s

b) Maximum altitude = 540 m

c) Range = 2160.6 m

d) r = (1440î + 480j) m

Explanation:

The distance at any time for the projectile is given by the relation - r² = x² + y²

where x = horizontal distance covered covered by the projectile and y = vertical distance coveredby the projectile

Let the initial velocity be u = ?

angle of projection be θ with respect to the horizontal = 45°

u = (uₓî + uᵧj) m/s

T = total time of flight = 21 s

t = any time during the flight of the projectile

a) Total time of flight = 2 uᵧ/g = (2u sin θ)/g

21 = (2u sin 45°)/9.8

u = 145.5 m/s

b) maximum altitude of the projectile = H

H = (u² sin² θ)/2g

H = (145.5² sin² 45°)/(2 × 9.8)

H = 540 m

c) According to projectile motion the maximum horizontal displacement is given by

x = R = uₓT = u cos(θ) T (since uₓ = u cos θ)

R = (145.5 cos 45°) × 21 = 2160.6 m

d) At 14 s,

x = uₓt = u cos(θ) t (since uₓ = u cos θ)

x = (145.5 cos 45°) × 14 = 1440.1 m

y = uᵧ t - 0.5gt² = [u sin(θ)] t - 0.5gt² = (145.5 sin 45°) × 14 - 0.5(9.8)(14) = 480 m

r = (1440î + 480j) m

6 0
2 years ago
On a dry day, just after washing your hair to remove natural oils and drying it thoroughly, run a plastic comb through it. Small
Hitman42 [59]
When the surface of the comb rubs on your hair, the comb is electrically charged. When the comb comes close to the paper, the charge on the comb causes charge separation on the paper bits. Since paper is neutral, positive and negative charges are equivalent. The charge on the comb charges the area of the bit of paper nearest the comb to the opposite. Thus, the bits of paper become attracted to the comb.
5 0
1 year ago
Light from a distant star is collected by a concave mirror that has a radius of curvature of 150 cm. How far from the mirror is
andrey2020 [161]
The focal point of a mirror is half of the radius of curvature. We can use the formula,                                                                                                                   1/f = 1/v + 1/u                                                                                                          where f is the focal length , v is the image distance and u is object distance        The distance of the star is assumed to be incredibly far away and 1 divided by a really big number is approximately zero, thus;                                                     1/f = 1/v  = 1/75 = 1/v                                                                                             therefore; the image is formed 75 cm infront of the mirror
4 0
2 years ago
Jack's model includes one object consisting of two molecules and one object consisting of six molecules. He put 16 energy cubes
Novay_Z [31]

Answer:

the molecules have different energy and the system is not in equilibrium

Explanation:

The model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.

 So when you book them together the molecules have different energy and the system is not in equilibrium

5 0
2 years ago
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