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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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In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
denis23 [38]
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
2 years ago
Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

N = 7.74 rev

Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

so we have

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)

\alpha = 0.48 rad/s^2

Part e)

now for uniform acceleration we have

\omega_f - \omega_i = \alpha t

12.56 - 8.17 = 0.48 t

t = 9.14 s

7 0
2 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
2 years ago
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