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3 years ago

The amount of energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin is called its

2 answers:

8 0

Specific heat. The definition of specific heat is the amount of energy required to raise the temperature of 1g of a substance by 1K or 1°C.

Elden [556K]3 years ago

3 0

Answer:

Specific heat

Explanation:

Specific heat capacity is the amount of energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin. According to the formula for heat capacity,(H)

H = mc∆t where m is the mass of the object, c is the specific heat capacity and ∆t is the change in temperature.

Specific heat capacity (c) from the formula will be c = H/m∆t

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How does the frequency of infrared electromagnetic waves compare with the frequency of radio and microwaves?

iVinArrow [24]

Answer:

Answer is B.

Because the wavelength of infrared is shorter than microwave radiation

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3 years ago

Why is air considered a fluid

34kurt

Answer:

A fluids is any substance that flows. Air is made of stuff, air particles, that are loosely held together in a gas form. Although liquids are the most commonly recognized fluids, gasses are also fluids. Since air is a gas, it flows and takes the form of its container.

Explanation:

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3 years ago

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

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3 years ago

How does 3rd class lever make our work easier

sineoko [7]

In a third class lever, the effort is located between the load and the fulcrum. If the fulcrum is closer to the load, then less effort is needed to move the load. If the fulcrum is closer to the effort, then the load will move a greater distance. ... These levers are useful for making precise movements.

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2 years ago

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The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul

ozzi

The correct answer is
B It increases.

In fact, the kinetic energy of a moving object is given by
$K= \frac{1}{2}mv^2$
where m is the mass of the object and v is its speed. We see that the kinetic energy is proportional to the mass and proportional to the square of the speed: in this problem, the speed of the object remains the same, while its mass increases, therefore the kinetic energy will increase as well.

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3 years ago

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