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3 years ago

Is molded bread a physical or chemical change?

1 answer:

8 0

<span>A molded bread is a good example of a chemical change. A chemical change is a kind of change in which a new substance is formed. In a chemical change, the original substance which started the reaction can not be recovered.The molded bread formed in this given instance is a new substance which is different from the original bread, so also, the original bread can no longer be retrieved from the molded bread.</span>

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How many neutrons are present in an atom of Iron-54

Flura [38]

there are 28 neutrons present

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3 years ago

A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m

charle [14.2K]

Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D) $x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

$K_{eq} + U_{eq} = K_A + U_A\\\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\v = \sqrt{\frac{k}{m}}A$

where $\omega = \sqrt{k/m}$ and $\omega = 2\pi f$

Therefore,

$v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s$

Part B

The conservation of energy will be used again.

$K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\mv^2 + kx^2 = kA^2\\(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\v = \sqrt{0.054k}$

where $k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89$

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

$E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J$

Part D

The general equation of motion in simple harmonic motion is

$x(t) = A\cos(\omega t + \phi)\\x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)$

where $\phi$ is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,

$x(t) = (0.15m)\cos(2\pi (2.0Hz)t)$

5 0

4 years ago

The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the

ycow [4]

$F = 5.93×10^{13}\:\text{N}$

Explanation:

Given:

$m_1= 2×10^{16}\:\text{kg}$

$m_2= 4×10^{22}\:\text{kg}$

$r = 30000\:\text{km} = 3×10^7\:\text{m}$

Using Newton's universal law of gravitation, we can write

$F = G\dfrac{m_1m_2}{r^2}$

$\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}$

$\:\:\:\:= 5.93×10^{13}\:\text{N}$

3 0

3 years ago

If you have a block of wood that is 4cm wide ,6cm long and 10cm high and it weighs 100 grams what would happen to the volume if

n200080 [17]

It would make both half 4cm to 2cm wide, and 6cm to 3cm long, 10cm to 5cm hight it would split in half

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3 years ago

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If the radius of orbit of a satellite is increased then the orbital speed would

JulsSmile [24]

Not be as fast as if the radius were smaller. Why? Because the closer you are to the magnetic field, the stronger the gravitational pull on the object would be.

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4 years ago