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4 years ago

True of false efficiency compared the output work to the output force

1 answer:

8 0

The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>

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During a single-replacement reaction, a more-active metal will replace a less-active metal in a compound. Which single-replaceme

son4ous [18]

Answer:

Mg will replace Ag in a compound

Explanation:

A single replacement reaction is driven by the position of ions on the activity series.

As a rule of thumb, the position of metal ions on the activity series determines their reactivity.

Metal ions that are above another are more reactive and they will displace those that are lower.

Generally, activity increases as we go up the group.

Mg ions are higher than Ag ions on the series so, Mg will displace Ag from a solution.

8 0

3 years ago

How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

stealth61 [152]

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

$W = P\Delta V$

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

$W = P(0\ m^3)\\$

5 0

3 years ago

3 In a television tube, an electron starting from rest experiences a force of 4.0 × 10−15 N over a distance of 50 cm. The final

MAXImum [283]

Answer:

The final speed of the electron = 2.095×10⁸ m/s

Explanation:

From newton's fundamental equation of dynamics,

F = ma ........................Equation 1

Where F = force, m = mass of the electron, a = acceleration of the electron.

making a the subject of the equation,

a = F/m.................... Equation 2

Given: F = 4.0×10⁻¹⁵ N,

Constant: m = 9.109×10⁻³¹ kg.

Substituting into equation 2

a = 4.0×10⁻¹⁵/9.109×10⁻³¹

a = 4.39×10¹⁶ m/s².

Using newton's equation of motion,

v² = u²+2as .......................... Equation 3

Where v = final velocity of the electron, u = initial velocity of the electron, a = acceleration of the electron, s = distance covered by the electron.

Given: u = 0 m/s(at rest), s = 50 cm = 0.5 m, a = 4.39×10¹⁶ m/s²

Substituting into equation 3

v² = 0² + 2(0.5)(4.39×10¹⁶)

v = √(4.39×10¹⁶)

v = 2.095×10⁸ m/s

Thus the final speed of the electron = 2.095×10⁸ m/s

7 0

3 years ago

A stereo speaker produces a pure \"c\" tone, with a frequency of 262 hz. what is the period of the sound wave produced by the sp

jasenka [17]

Answer:

3.82 ms

Explanation:

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f is the frequency.

In this problem, f = 262 Hz, so the period if this sound wave is

$T=\frac{1}{262 Hz}=3.82\cdot 10^{-3} s=3.82ms$

4 0

3 years ago

What is the hottest layer in the Sun's atmosphere?

erica [24]

I believe it is the core

7 0

3 years ago