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2 years ago

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Longitudional waves travel through a series of and ___.

1 answer:

nekit [7.7K]2 years ago

4 0

Answer:

compressions; rarefactions

Explanation:

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solids, liquids, and gases are three forms of matter that:a. take up space. b. have mass. c. are made of atoms. d. all of the ab

LiRa [457]

Answer:

d

Explanation:

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2 years ago

What phenomenon causes colors of visible light to be separated by a prism?

Allushta [10]

Dispersion im pretty sure

6 0

2 years ago

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

3 years ago

An astronaut weighing 248 lbs on Earth is on a mission to the Moon and Mars. (a) What would he weigh in newtons when he is on th

lesantik [10]

Answer:

$W_m=183.495N$

Explanation:

From the question we are told that:

Weight $W=248Ibs$

Mass of Weight $m= 248*0.453$

$m=112.344kg$

Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.

Therefore

The equation for Weight on Moon is given as

$W_m=m*g/6$

$W_m=122.344*\frac{9.8}{6}$

$W_m=183.495N$

3 0

2 years ago

Read 2 more answers

Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0°C whe

m_a_m_a [10]

Answer: The ice absorb 6671.1 kJ of thermal energy.

Explanation:

The conversions involved in this process are :

$0.00^0C=273K$

$:H_2O(s)(273K)\rightarrow H_2O(l)(273K)$

Now we have to calculate the enthalpy change.

$\Delta H=n\times \Delta H_{fusion}$

where,

$\Delta H$ = enthalpy change = ?

m = mass of ice = 20.0 kg = $20.0\times 10^3g$ (1kg=1000g)

n = number of moles of ice= $\frac{\text{Mass of ice}}{\text{Molar mass of water}}=\frac{20.0\times 10^3g}{18g/mole}=1.11\times 10^3mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=1.11\times 10^3mole\times 6010J/mole$

$\Delta H=6671100J=6671.1kJ$ (1 kJ = 1000 J)

Therefore, the enthalpy change is, 6671.1 kJ

6 0

2 years ago