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Dmitriy789 [7]
3 years ago
8

. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T

he first vehicle collides with a concrete barrier, and stops in a time of approximately 0.1 s. The second vehicle collides with a collapsible barrier, and comes to rest in about 1 second. Which object is subject to a larger force
Physics
1 answer:
Serggg [28]3 years ago
5 0

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

         I = F t = Δp

in this case the final velocity is zero

        F t = 0 -m v₀

        F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

        F₁ = -p₀ / 0.1

        F₁ = - 10 p₀

barrier collapses

         F₂ = -p₀ / 1

let's look for the relationship of the forces

        F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

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Explanation:

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g=3n

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If we know the restaurant will make 8 pizzas (n=8), then:

g=3(8)

g=24 This is the needed number of mushrooms for 8 pizzas

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3 years ago
Calculate the current through a 3.0 ω resistor with a voltage of 9.0 v across it.
gulaghasi [49]

Answer: 3 A

Explanation:

According to<u> Ohm's law</u>:  

V=R.I

Where:

V=9 V is the voltage

R=3\Omega is the resistance of the resistor

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I=\frac{V}{R}

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7 0
3 years ago
A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine
Goryan [66]

Answer:

a) 2.85 kW

b) $ 432

c) $ 76.95

Explanation:

Average price of electricity = 1 $/40 MJ

Q = 20 kW

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also

K=(QH)/Win

Now,

Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7

where

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Thus,

P(in) = 20/7 = 2.85 kW

b) Cost = Energy consumed × charges

Cost = ($1/40000kWh) × (16kW × 300 × 3600s)

cost = $ 432

c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95

4 0
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6 0
3 years ago
Read 2 more answers
Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep
SIZIF [17.4K]

Answer:

1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

              f = r / 2

              f = 2/2

             f = 1 m

The builder's equation

           1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

              1 / f = 0 + 1 / i

              i = f

              i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

              R₂ = infinity

              R₁ > 0

Cas2 Flat-concave (divergent) lens

             R₂ = infinity

              R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

          1 / f₂ = 1 / (L-f₁) + 1 / i

          1 / i = 1 / f₂ - 1 / (L-f₁)

           1 / i = (L-f₁-f₂) / f₂ (L-f₁)

           i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0
3 years ago
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