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Zina [86]
3 years ago
10

A 45.0-kg person steps on a scale in an elevator. The scale reads 460 n. What is the magnitude of the acceleration of the elevat

or?
Physics
1 answer:
S_A_V [24]3 years ago
7 0

Answer:

0.42 m/s^2

Explanation:

There are two forces acting on the person as he's standing on the scale in the elevator:

- Its weight, mg, downward, with m = 45.0 kg being his mass and g = 9.8 m/s^2 being the acceleration of gravity

- The normal reaction exerted by the scale on the person, N, upward, which is equal to the value read on the scale, N = 460 N

Using Newton's second law, we can write

N-mg = ma

Where a is the acceleration of the person (and of the elevator). Solving for a,

a=\frac{N-mg}{m}=\frac{460-(45)(9.8)}{45}=0.42 m/s^2

And since we chose positive as the upward direction, the acceleration is upward.

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Artemon [7]

Answer:

D

Explanation:

Scientists use significant figures to avoid claiming more accuracy in a calculation than they actually know.

6 0
3 years ago
Read 2 more answers
I will Mark Brainliest 1. ) Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is
WITCHER [35]
The answer to question one is A.
The answer to question two is A.
The answer to question three is D.

7 0
3 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the tot
zheka24 [161]

The portion of the flux leaves the curved surface of the cylinder is 60%.

<h3 /><h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.

If 20% of the flux leave from one end, then another 20% will leave from another end.

So, the net flux through curved surface is

100 -20 -20 = 60%

Thus, the total flux  leaves the curved surface of the cylinder is 60%

Learn more about electrons.

brainly.com/question/1255220

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5 0
1 year ago
If a bean of mass 2.0g jumps 1.0cm from your hand into the air, how much potential energy has it gained in reaching its highest
kirill [66]
PE = mg\Delta h = 0.002 \, kg \cdot 9.8 \, m/s^2 \cdot 0.01 \, m = ~2 \cdot 10^{-4} \, J
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