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lapo4ka [179]
3 years ago
13

A steel rope is used to lift a 26 kg slab of concrete from the ground to a height of 20 m. Assume that the rope moves the concre

te at a constant speed and then stops once the slab is at the final position above the ground. How much work does the steel rope do on the slab of concrete?
Physics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

5200 Joules

Explanation:

Work Formula:

W = F . D

W = (26.10) . 20

W = 260 . 20

W = 5200 Joules

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Joanna learned from customer satisfaction surveys that diners in her restaurant wanted to be able to make substitutions. She has
Dmitry [639]

Answer:

standards

Explanation:

Based on the information provided within the question in regards to the situation at hand it can be said that Joanna is demonstrating a standards gap. this is a gap caused by the difference between the customer service standards a company has created for itself and the expectations the company believes that the customers have for that company. Since Joanna did not tell all the servers of the customers expectations then the ones who do not know will not be able to provide this service to those customers, thus the restaurant will not be able to meet it's customer service standards.

3 0
3 years ago
Please answer this.
lilavasa [31]

Answer:

the answer is c

Explanation:

it's c because the moon has to be a full moon to be a solar eclipse when the sun moon and earth line up

6 0
3 years ago
Read 2 more answers
Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg
nata0808 [166]

Answer:

a) U_{e} = 3 \times 10^{10}\,J, b) v \approx 7745.967\,\frac{m}{s}

Explanation:

a) The potential energy is:

U_{e} = Q \cdot \Delta V

U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)

U_{e} = 3 \times 10^{10}\,J

b) Maximum final speed:

U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }

The final speed is:

v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }

v \approx 7745.967\,\frac{m}{s}

3 0
3 years ago
A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th
Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m

6 0
3 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
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