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3 years ago

What state of matter can easily be compressed (squeezed into a much smaller space)? *

Physics

1 answer:

avanturin [10]3 years ago

6 0

Answer:

gases.

Explanation:

gases have the highest volume. if you compress (increase pressure) a gas, the volume decreases and it will take up less space as a liquid, and if you compress more, the liquid will become a solid.

additionally info: this is generally true, but as you get into more difficult subjects, this rule does not hold. for example, some substances cannot obtain a certain phase of matter when they are a certain temperature and or pressure. but that could be way more out of this questions league.

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A street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar

____ [38]

Answer:

so you take 120÷2 wires

4 0

3 years ago

A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex

Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

$\text{Efficiency of the heat engine} = \eta = 26\% = 0.26$

$\text{Energy taken in by the heat engine during each cycle} = Q_h$

$\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J$

$\eta = 1 - \frac{Q_{c}}{Q_{h}}$

$0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}$

$\frac{8.55* 10^{3}}{Q_{h}} = 0.74$

$Q_h = \frac{8.55*10^3}{0.74}$

$Q_h = 11.554*10^3J$

PART A)

Work done by the heat engine in each cycle = W

$W = Q_h-Q_c$

$W = 11.554*10^3J-8.55*10^3J$

$W = 3004J$

According to the value given we have that,

$P = 4.0kW$

$P = 4000W$

Power is defined as the variation of energy as a function of time therefore,

$P = \frac{W}{t}$

$4000W = \frac{3004J}{t}$

$t = \frac{3004}{4000}$

$t = 0.75s$

Therefore the interval for each cycle is 0.75s

5 0

3 years ago

It takes 1 minute for 45 c to pass a point in a circuit, what is the current flowing through the circuit?

ohaa [14]

Answer:

0.75 A

Explanation:

An electric current is a flow of charged particles.

A current is defined through its intensity, which is given by:

$I=\frac{q}{t}$

where

I is the current intensity

q is the charge passing through a given point in the circuit

t is the time it takes for the charge q to pass that point in the circuit

In this problem, we have:

q = 45 C is the charge passing through the point in the circuit

$t=1 min = 60 s$ is the time elapsed

Therefore, the current intensity is:

$I=\frac{45}{60}=0.75 A$

5 0

3 years ago

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi

jeka94

Answer:

I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

d = (d₁ + d₂) / 2

d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

M = M_man - 0.2 M_man = 80 (1-0.2)

M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis: Ic = ½ M r²

A rod that rotates at the end: I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

I₁ = 8.258 + 5.33 + 4.129

I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

I₂ = ½ M r² + 2 m D²

I₂ = ½ 64 0.508² + 2 8 0.508²

I₂ = 8,258 + 4,129

I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

I₁ / I₂ = 17,717 /12,387

I₁ / I₂ = 1.43

3 0

3 years ago

New theories usually don’t start “from scratch.” For example, the modern model of the atom is the result of new information adde

Evgen [1.6K]

Answer:

Explanation:

7 0

3 years ago