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2 years ago

15

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa

ter is 4.186×10 to the third power J/kg times degrees Celsius

1 answer:

Anna [14]2 years ago

4 0

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature $T_{1}$ = 68 °F = 20° c

Final temperature $T_{2}$ = 212 °F = 100° c

Specific heat of water $C = 4.186 \frac{KJ}{kg c}$

Now heat transfer Q = m × C × ( $T_{2}$ - $T_{1}$ )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

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A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt

Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m , λ₂ = 2.175 10⁻¹⁰ m, E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

Ef = E₁ + E₂

E₀ = Ef

E₀ = E₁ + E₂

The energy can be found with the Planck equation

E = h f

c = λ f

f = c / λ

E = hc / λ

They indicate that the wavelength of the second photon is

λ₂ = λ₁ +0.130 10⁻⁹

We replace

E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ + λ₁) / λ₁ ( λ₁ + 0.13 10⁻⁹)

1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹ λ₁) = 2 λ₁ + 0.13 10⁻⁹

λ₁² + 0.13 10⁻⁹ λ₁ = 1.25 10⁻¹⁰ λ₁ + 8.125 10⁻²¹

λ₁² + 0.005 10⁻⁹ λ₁ = 8.125 10⁻²¹

λ₁² + 5 10⁻¹² λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

λ₁ = 87.5 10⁻¹² m

λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

λ₁ = 87.5 10⁻¹² m

λ₂ = λ₁ + 0.13 10⁻⁹

λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

E₂ = E₀ –E₁

E₂ = E₀ –hc / λ₁

E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

E₂ = 0.93 10⁻¹⁵ J

E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

E₂ = 5.8 10³ eV

7 0

3 years ago

A transport truck pulls on a trailer with a force of 600N [E]. The trailer pulls on the transport truck with a force

kramer

These forces form a force pair. Use Newton's third law, and you see that the trailer pulls back at with the same force. The answer is d.

6 0

3 years ago

Read 2 more answers

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

lesantik [10]

Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

5 0

2 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple

Finger [1]

Kepler derived his three laws of planetary motion entirely from
observations of the planets and their motions in the sky.

Newton published his law of universal gravitation almost a hundred
years later. Using some calculus and some analytic geometry, which
any serious sophomore in an engineering college should be able to do,
it can be shown that IF Newton's law of gravitation is correct, then it MUST
lead to Kepler's laws. Gravity, as Newton described it, must make the planets
in their orbits behave exactly as they do.

This demonstration is a tremendous boost for the work of both Kepler
and Newton.

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3 years ago

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