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3 years ago

12

1) Consider the data table as well as the experimental design. Which of the four cans should have the greatest change in tempera

ture after five minutes?
A) air
B) sand
C) water
D) iron filings

1 answer:

aleksandrvk [35]3 years ago

3 0

Answer:

Explanation:

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Mrac [35]

Answer:

It’s b

Explanation:

Everyone else is wrong and I tried this and got it right

6 0

3 years ago

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Can someone help me with #7

bekas [8.4K]

The ratio of (new volume / old volume) is the same as the ratio of (new temperature / old temp).

Just remember that you have to use the ABSOLUTE temperatures.

An absolute temperature is (Celsius + 273).

7 0

4 years ago

The diagram below shows different layers of sedimentary rocks.

Arlecino [84]

Layer F was formed earlier than Layer A

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3 years ago

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Two identical waves are moving in the same direction with the same speed. If the amplitude of the combination of the two waves i

umka2103 [35]

Answer:

The phase difference between these two waves is 141.1⁰

Explanation:

The displacement of the wave is given as;

$Y = y_xSin(Kx - \omega t)+y_xSin(Kx- \omega t + \phi)\\\\Y = 2y_xCos(\frac{1}{2} \phi)Sine(Kx- \omega t + \frac{1}{2} \phi)$

Amplitude, A = 2yₓCos(¹/₂Φ)

Since the amplitude of the combination is 1.5 times that of one of the original amplitudes = yₓ = 1.5 × A = 1.5A

A = 2(1.5A)Cos(¹/₂Φ)

A = 3ACos(¹/₂Φ)

¹/₃ = Cos(¹/₂Φ)

(¹/₂Φ) = Cos ⁻(0.3333)

(¹/₂Φ) = 70.55°

Φ = 141.1°

The phase difference between these two waves is 141.1⁰

4 0

4 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago