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4 years ago

How much heat is required to warm 1.40 l of water from 20.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the water.)?

1 answer:

3 0

The water density is
$d=1.0 g/mL = 1000 g/L$
And the mass of 1.40 L of water is
$m=dV=(1000 g/L)(1.40 L)=1400 g$

The amount of heat needed to increase the temperature of the water by $\Delta T$ is given by
$Q=m C_s \Delta T$
where m is the water mass, $C_s = 4.18 J/g ^{\circ}C$ is the water specific heat capacity and
$\Delta T=100.0 ^{\circ}C-20.0 ^{\circ}C = 80.0^{\circ}C$
is the increase in temperature. If we substitute these numbers into the equation, we find
$Q=(1400 g)(4.18 J/g^{\circ}C)(80.0^{\circ}C)=4.68 \cdot 10^5 J$

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Two coils, held in fixed positions, have a mutual inductance of M = 0.0034 H. The current in the first coil is I(t) = I0sin(ωt),

tiny-mole [99]

Answer:

ε₂ =2.63 V

Explanation:

given,

M = 0.0034 H

I (t) = I₀ sin (ωt)

I (t) = 5.4 sin (143 t)

$\dfrac{d i(t)}{dt} = \dfrac{d}{dt}(5.4 sin (143 t))$

$\dfrac{d i(t)}{dt} =143 \times 5.4 cos (143 t)$

magnitude of the induced emf in the second coil

ε₂ = $M\dfrac{di}{dt}$

ε₂ = $0.0034\times 143 \times 5.4 cos (143 t)$

for maximum emf

cos (143 t) = 1

ε₂ = $0.0034\times 143 \times 5.4$

ε₂ =2.63 V

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3 years ago

Charlie graduated with a bachelor’s degree in construction management. As he aspires to work in his dad’s construction firm, whi

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Answer:

He needs to enroll himself in CMMA. This company help young students obtain a well future in Construction Science.

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3 years ago

What is the wavelength of the sound wave that reaches your ear in a 20∘c room?

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Missing text in the problem (found on internet):
"<span>A 70 cm -long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamental frequency"

Solution:
The frequency of the sound wave along the string is given by
</span> $f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
<span>where L=70 cm=0.7 m is the length of the string, T=200 N its tension and </span> $\mu = 1g/m =0.001 kg/m$ is the linear density. Using these values in the formula, we find
$f=319.4 Hz$

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3 years ago

Emissions from alpha decay require the least amount of shielding. Which best describes why the emissions from alpha decay are so

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Answer:

Alpha decay releases particles that have more mass and charge than the particles released during beta decay.

Explanation:

In the alpha decay, an alpha particle is emitted. An alpha particle is a nucleus of helium, consisting of two protons and two neutrons. The particle emitted in the beta decay is instead an electron: therefore, the alpha particle has more mass and more charge than a beta particle. For this reason, alpha particles are much more ionising than electrons, so they release much more energy per unit distance than beta particles, and so they lose all their energy much faster than beta particles. This is why alpha particles can be easily blocked even by a thin piece of paper, while beta particles are able to pass through.

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3 years ago

Read 2 more answers

For 1983 through 1989, the per capita con-

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Answer:

y = 2.2t + 30.2

Explanation:

t = 3 represents 1983, and t = 9 represents 1989.

Two points on the line are (3, 36.8) and (9, 50).

First, find the slope of the line.

m = (50 − 36.8) / (9 − 3)

m = 2.2

Now use point-slope form to write the equation:

y − 36.8 = 2.2 (t − 3)

If you wish, simplify to slope-intercept form:

y − 36.8 = 2.2t − 6.6

y = 2.2t + 30.2

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3 years ago