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3 years ago

Use the drop-down menus to identify each labeled part of the diagram.

Physics

1 answer:

iren2701 [21]3 years ago

3 0

Answer:

<u>A:cool fluid sinks</u>

<u>B:warm fluid rises</u>

<u>C:convection current</u>

Explanation:

Just took the assessment!!

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A sports car engine delivers 150 kW to the driveshaft with a thermal efficiency of 20%. The fuel has a heating value of 35 000 k

Andreyy89

Answer:600 kW

Explanation:

Given

Power delivered $P=150 kW$

Fuel heating value $q=35,000 kJ/kg$

thermal Efficiency $\eta =\frac{Power}{Heat}$

$0.2=\frac{150}{35,000\times \dot{m}}$

where, $\dot{m} =mass\ flow\ rate\ of\ fuel$

$\dot{m}=\frac{150}{35,000\times 0.2}$

$\dot{m}=0.0214 kg/s$

For an Engine

$q\dot{m}=P+Q_r , where Q_r=heat\ rejected$

$Q_r=q\dot{m}-P$

$Q_r=750-150=600 kW$

3 0

3 years ago

John rode 3,150 m at an average speed of 350 m/min. If he

Vikentia [17]

Answer:

mabye 8

Explanation:

6 0

3 years ago

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

gayaneshka [121]

Answer:

The value is $E = 1.35 *10^{14} \ J$

Explanation:

From the question we are told that

The mass of matter converted to energy on first test is $m = 1 \ g = 0.001 \ kg$

The mass of matter converted to energy on second test $m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg$

Generally the amount of energy that was released by the explosion is mathematically represented as

$E = m * c^2$

=> $E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2$

=> $E = 1.35 *10^{14} \ J$

7 0

3 years ago

A girl on roller skates accelerates at a rate of 2 m/sec/sec with a force of 100 N. What is her mass?

Dvinal [7]

By the newtons Second Law:

F = ma

Solving for m:

m = F / a

m = 100 N / 2 m/s²

<h3>m = 50 kg → ANSWER</h3>

8 0

3 years ago

The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

6 0

4 years ago