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2 years ago

9

Một khối khí hidro bị nén đến thể tích bằng 1/2 lúc đầu khi nhiệt độ không đổi. Nếu vận tốc trung bình của phân tử hidro lúc đầu

là V thì vận tốc trung bình sau khi nén là bao nhiêu ?

1 answer:

BartSMP [9]2 years ago

7 0

answer: what language is this???

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Sam is pulling a box up to the second story of his apartment via a string. The box weighs 53.3 kg and starts from rest on the gr

castortr0y [4]

Answer:

W = 1222.4 J = 1.22 KJ

Explanation:

The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:

W = F d Cosθ

where,

W = Work Done = ?

F = Force Applied = 64 N

d = Distance Covered by Box = 19.1 m

θ = Angle between force and displacement = 0°

Therefore,

W = (64 N)(19.1 m)Cos 0°

<u>W = 1222.4 J = 1.22 KJ</u>

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2 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca

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Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. $s_{o}$ = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

$v_{i}$ = 33.2 m/s, a = 0 (since the velocity is constant), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

s = 33.2t .......... eq (1)

Hatchback:

$a=5m/s^{2}$ , $v_{i}$ = 0 m/s (since initial velocity is zero), $s_{o}$ = 0

Using $s =s_{o}+v_{i}t+1/2at^{2}$

putting in the data we will get

$s=(1/2)(5)t^{2}$

now putting 's' value from eq (1)

$2.5t^{2}-33.2t=0$

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

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3 years ago

Read 2 more answers

A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260

ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

$\tau=F\times r$

$I\times\alpha=F\times r$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r$

$F=\dfrac{M}{2}a$ ...(I)

Here, F = tension

The force is

$F=m(g-a)$ ...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

$\dfrac{M}{2}a=m(g-a)$

$a=\dfrac{g}{1+\dfrac{M}{2m}}$

Put the value into the formula

$a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}$

$a=6.84\ m/s^2$

We need to calculate the tension in the rope

Using equation (I)

$F=\dfrac{M}{2}a$

Put the value into the formula

$F=\dfrac{12.1}{2}\times6.84$

$F=41.38\ N$

Hence, The tension in the rope is 41.38 N.

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2 years ago

In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

Aleonysh [2.5K]

Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light $I$ = 2500 W/m^2

speed of light $c$ <u> </u>= 3 x 10^8 m/s

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b) 1 atm = 101325 Pa

$P_{abs}$ = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

$P_{ref}$ = $2I/c$ = twice the value for totally absorbing

$P_{ref}$ = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d) 1 atm = 101325 Pa

$P_{ref}$ = 2 x 8.23 x 10^-11 = <em>1.65 x 10^-10 atm</em>

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2 years ago

What happens when a mixture is separated

iris [78.8K]

Answer:

the answer would be c

Explanation:

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2 years ago