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2 years ago

Heat engines use heat to do work. true or false.

2 answers:

7 0

Heat engines transform heat energy to other type of energy, typically mechanical or electrical, so yeah, they use heat to work.

ankoles [38]2 years ago

4 0

True.

Heat engines makes use of heat to do work.

You might be interested in

Does A= 6.4 B=12 C=12.2 Form a RIGHT TRIANLGE?

Lina20 [59]

To answer this question, you must remember the equation:

a²+b²= c²

(6.4)² + (12)²= (12.2)²

<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84

This cannot be a triangle

hope this helps

3 0

3 years ago

in a railyard a train is being put together from freight cars.An empty freight car coasting 10 m/s strikes a loaded car that is

svetoff [14.1K]

Answer:m1v1 + m2v2 = (m1f + m2f)vf. 3000kg(10.0m/s) + (15000kg)(0.0m/s) = (18000kg)(vf).

Explanation:

7 0

2 years ago

Convert 13.1 miles to feet. Using one step conversion

Marat540 [252]

Answer:

69,168 ft

Explanation:

6 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon

denpristay [2]

Answer:

For C1, Q = 1.6125×10⁻³ C

For C2, Q = 6.25×10⁻⁴ C

Explanation:

Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.

From the question,

Q = CV........................ Equation 1

Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.

For the first capacitor,

Q = C1V............. Equation 2

Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

Substitute into equation 2

Q = (6.45×10⁻⁶ )(250)

Q = 1.6125×10⁻³ C.

For the the second capacitor,

Q = C2V............. Equation 3

Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V

Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

4 0

3 years ago