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2 years ago

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The new horizons spacecraft, launched in 2006, spent 9.5 years on its journey to pluto. it derives its electric power from the h

eat generated by 10 kg of , which has an activity of . each decay emits an alpha particle with an energy of 5.6 mev. what is the total thermal power generated by this plutonium source

1 answer:

natta225 [31]2 years ago

7 0

The total thermal power generated by this plutonium source is determined as 5.65 J/s.

<h3>Total thermal power generated</h3>

The total thermal power generated by this plutonium source is determined as follows;

ΔP = Eλ

where;

E is the energy of the particles = 5.6 mev
λ is the activity of the plutonium, which should be = 6.3 x 10¹⁵ s⁻¹

ΔP = (5.6 mev) x (6.3 x 10¹⁵ s⁻¹)

ΔP = (5.6 x 10³ x 1.6 x 10⁻¹⁹ J) x (6.3 x 10¹⁵ s⁻¹)

ΔP = 5.65 J/s

Learn more about thermal power here: brainly.com/question/7541718

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4 years ago

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t

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Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

$q_{h}$ = R* $T_{h}$ in( $V_{2}$ / $V_{1}$ )

$=0.287 * 1200 ln(3)$

$=0.287*1318.33$

$=378.36kJ/kg$

The specific heat capacity:

$q_{L}$ =q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

$n_{TH}$ =1 - ( $T_{L}$ / $T_{H}$ )

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

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3 years ago

Which type of radiation travels at the speed of light and penetrates matter easily?

Alina [70]

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3 years ago

A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick

Rom4ik [11]

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0.5mv²
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3 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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4 years ago