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2 years ago

When melted does the size of butter change

Physics

2 answers:

patriot [66]2 years ago

6 0

The size changes but the mass does not

bixtya [17]2 years ago

4 0

No. Because the solid and liquid is in the same volume

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What is the equation used to calculate the total amount of energy used by an appliance?

charle [14.2K]

Answer:

Power = Current × Voltage

Explanation:

Units:

Power = Watts

Current = Àmperes

Voltage = Volts

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2 years ago

Two sounds are played for you. The first has a frequency of 196 Hz. The second has a frequency of 392 Hz. Which sound will be "h

kykrilka [37]

Answer:

The second one, the higher the hertz the higher the frequency we hear

Explanation:

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2 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

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2 years ago

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Jet001 [13]

1.b 2.c 3.a 4.b 5.d 6.a 7.a 9.idk 10.c hope thes helps have a nice day and God bless

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2 years ago

For a steady two-dimensional flow, identify the boundary layer approximations.

Georgia [21]

Answer:

The velocity component in the flow direction is much larger than that in the normal direction ( A )
The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )

Explanation:

For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction

assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that

Vy << Vx

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2 years ago