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Luden [163]
2 years ago
10

A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters,

and its height is 0.85 meters. What is the percentage efficiency if a force of 496 newtons is applied to the refrigerator?
A. 44 percent
B. 58 percent
C. 64 percent
D. 76 percent
E. 92 percent
Physics
2 answers:
irinina [24]2 years ago
7 0
The answer will be D. 76 percent since 496 is out of 1,127N
lina2011 [118]2 years ago
3 0
496/1127 = 0.44 = 44% 

<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>

<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>

<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>

<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>
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2 years ago
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A mercury atom in the ground state absorbs 20.00 electronvolts of energy and is ionized by losing an electron. How much kinetic
sukhopar [10]
We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.

The initial energy is 20 eV = 20 J/C
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To determine the kinetic energy, we can use this equation:

KE = 20 Joules / Coulombs * </span>1.60217662 × 10-19<span> coulombs 
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8 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w
Artemon [7]

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

7 0
2 years ago
Two parallel co-axial disks are floating in deep space (far from sun and planets). Each disk is 1 meter in diameter and the disk
HACTEHA [7]

Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

          P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of ​​the disk, T the absolute temperature and e the emissivity that for a black body is  1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

           I = P / A

Writing this expression for both discs

          I₁ A₁ = I₂ A₂

          I₂ = I₁ A₁ / A₂

The area of ​​a sphere is

          A = 4π r²

           I₂ = I₁ (r₁ / r₂)²

          r₂ = r₁ ± 5

          I₁ = I₂ ( (r₁ ± 5)/r₁)²

.

        Let's write the Stefan equation

         P / A = σ e T⁴

          I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

         σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values ​​of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

        e₁ T₁⁴ = T₂⁴   (r1 + 5)²/r₁²

       T₁ = T₂  {(e₂/e₁)}^{1/4}  √(1 ± 1/ r₁)  

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

           (1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

          T₁ = T₂ {(e₂/e₁)}^{1/4}  (1 ± ½   5/r1)

           T₁ = T₂ {(e₂)}^{1/4}   (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

              T₁ = T₂  {(e₂)}^{1/4} ¼

              T<u>₁</u> = T₂  0.90.9^{1/4}

               5500 = T₂  0.974

               T₂ = 5646 K

3 0
2 years ago
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cupoosta [38]
Thats hard...im not sure
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