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1 year ago

Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-

Physics

1 answer:

Nady [450]1 year ago

7 0

Answer:

a) W_total = 8240 J , b) W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

F - W = m a

W = mg

F = m (a + g)

F = 80 (1 + 9.8)

F = 864 N

The work of this force we will call it W1

We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

F₂ - W = 0

F₂ = W

F₂ = 80 9.8

F₂ = 784 N

The work of this fura we will call them W2

The total work is

W_total = W₁ + W₂

W_total = (F + F₂) y

W_total = (864 + 784) 5

W_total = 8240 J

b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

W₁ / W₂ = F y / F₂ y

W₁ / W₂ = 864/784

W₁ / W₂ = 1.1

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

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2 years ago

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) On doubling the mass:

New mass, $m'=2m$

Then the new time period:

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

Then the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

Then the new time period:

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

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1 year ago

The brain receives messages in signals called nerve impulses. Which part of the ear first generates these nerve impulses?

liberstina [14]

That's the job of the tiny "hair cells", located in the inner ear.

If you're a sound wave, this is how you reach the hair cells:

-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum"; sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear". In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate; then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear"; it's a long skinny chamber, called the "cochlea", wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.

-- there are two complete copies of this whole structure ... one on each side of your head.

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2 years ago

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sesenic [268]

Velocity =

(distance between start point and end point, regardless of the route traveled) / (time spent traveling).

That distance (called the "displacement"), is 10 meters, and almost exactly 1 hour is almost exactly 3,600 seconds. So the numerical value of the velocity during that time is

(10) / (3,600) = almost exactly 0.00278 m/s

= 2.78 x 10^-3 m/s.

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2 years ago

Why are we unable to work long without food

nataly862011 [7]

Answer:

Without food your not able to produce energy thats why you cannot work so long

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