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fiasKO [112]
3 years ago
11

An old lorry is 20% efficient at converting the

Physics
1 answer:
AleksandrR [38]3 years ago
4 0

Answer:

The amount of mechanical energy produced by the old lorry when 600 Joules of chemical energy have been used by the engine, M.E. is 120 Joules

Explanation:

The question is a word problem on the relationship between energy efficiency and energy consumption

The given parameters of the old lorry are;

The chemical energy to mechanical energy conversion efficiency of the old lorry = 20%

The amount of chemical energy used by the engine = 600 joules

Let 'M.E.' represent the amount of mechanical energy produced by the old lorry, we have;

Energy \ Efficiency  = \dfrac{Energy \ produced  }{Total \ Energy \ Taken \ In \ or \ Used} \times 100

Therefore, the for the old lorry, we have;

20\% = \dfrac{M.E. \ Produced}{600 \ Joules} \times 100

∴ M.E. Produced = 20/100 × 600 J = 120 J

The amount of mechanical energy produced by the old lorry when 600 Joules of chemical energy have been used by the engine, M.E. = 120 Joules

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2 years ago
What is the direction of the current flowing through the wire—left or right? The current flows to the .
muminat

Answer:

right

Explanation:

Current flows from the positive (+ve) terminal of the battery to the negative (-ve). This is called conventional current flow.

Therefore, the electrons are negatively charged and want to get away from the Negative Terminal and go to the Positive Terminal, Hence the electrons move from left to right and current flows from right to left.

5 0
3 years ago
Read 2 more answers
Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund
vova2212 [387]

Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

6 0
3 years ago
Light in vacuum travels at a speed of 3.00 x 10^8 m ^-1 s^-1 on average earth is 93,000,000 miles from the sun how many minutes
NemiM [27]
The correct unit for the speed of light is  [ m s⁻¹ ].

Time = (distance) / (speed)

Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .

That would be  <em>8.31 minutes</em>.  
6 0
2 years ago
During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0
madreJ [45]

Answer:

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4orbits}

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km/Orbit

Explanation:

Given Data:

Numbers of times Telescope cycled around the earth in 6 years=37,000 times

Total Distance traveled in 6 years by the Hubble Space Telescope=1,280,000,000 Km

Find:

Kilometers in one Orbit=?

Solution:

Kilometers in 37,000 Orbits=1,280,000,000 Km

Kilometers in 1 Orbit=1,280,000,000/37,000

In Scientific Notation:

Kilometers\ in\ 3.7*10^4\ Orbits=1.28*10^9 Km

Kilometers\ in\ 1\ Orbit=\frac{1.28*10^9 Km}{3.7*10^4 orbits}

Kilometers in 1 Orbit=34594.594 Km

Kilometers in 1 Orbit in Scientific notation:

Kilometers\ in\ 1\ Orbit=3.46*10^4 Km

8 0
3 years ago
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