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8_murik_8 [283]
2 years ago
3

Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en

tire structure would sink by a certain amount for every additional floor that is added. If the maximum permissible limit for depression of the structure is 20 centimeters, how many floors can be safely added to the building?
Physics
1 answer:
Arlecino [84]2 years ago
8 0
If the maximum permissible limit for depression of the structure is 20 centimeters, the number of floors that can be safely added to the building is C. 18

depression = (depression/floor)(# floors) < 20

Here are the following choices:
<span>A. 14
B. 15
C. 18
D. 23</span>
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A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

    The mass suspended in mid-air is m_a =  5.0 \mu g =  5.0 *10^{-6} \ g =  5.0 *10^{-9} \  kg

   

Generally the electric field on the carpet is mathematically represented as

           E =  \frac{q}{ 2 *  A  *  \epsilon _o}

Where \epsilon _o is the permittivity of free space with value \epsilon_o = 8.85*10^{-12}  \ \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

           E =  \frac{-10*10^{-6}}{ 2 *  (2 * 4 )  *  8.85*10^{-12}}

           E = -70621.5 \  N/C

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

=>     q_d  = 6.94 *10^{-13} \  C

4 0
2 years ago
The graph indicates Linda’s walk.
Sedaia [141]
I think the right answer is the first one. If she stops moving her Position does not change any more-and the Graph Shows that after 6 seconds she stays at the Position of 5 m. If she Went Back to the start point the Graph would have Developed Back to 0m(decreased).
3 0
2 years ago
Read 2 more answers
Juan was wearing a bright red shirt in a very dark room. What color did his shirt appear to the people with him in the room? A)
ikadub [295]
It would appear black.   

Hope I helped.  
8 0
2 years ago
Read 2 more answers
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The
Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

​

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0
2 years ago
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