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andrew-mc [135]
3 years ago
8

Suchita is going to make an energy transfer diagram to represent the energy transformations associated with the flame of a gas s

tove. The diagram will show that the chemical energy of the fuel transforms into 50 J of chemical energy, 400 J of heat, and 200 J of light.
Which statement describes Suchita’s final diagram?


The initial chemical energy is 50 J.

The arrow for heat is the widest output arrow.

The arrow for chemical energy is the shortest output arrow.

There is one arrow for input and one for output.
Physics
2 answers:
ira [324]3 years ago
7 0
The answer the the question is, B.<span />
katrin [286]3 years ago
7 0

Answer:

The arrow for heat is the widest output arrow.

The arrow for chemical energy is the shortest output arrow.

Explanation:

here by the law of energy conservation we can say that

Total energy that is used or consumed must be equal to the total energy that is transferred in different forms

So here we will say that

Q_{in} = Q_{out}

here output energies are given as

1. chemical energy = 50 J

2. Heat = 400 J

3. Light = 200 J

so here in output energy form we will get maximum energy in form of heat so the arrow related to this must be widest in size

While the chemical energy is minimum output energy so here the least size arrow will be of chemical energy

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Answer:

  a = 7.5 m / s²

Explanation:

For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane

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X axis

       W sin θ = m a

 

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let's calculate

        a = 9.8 cos 40

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8 0
2 years ago
Janelle stands on a balcony, two stories above Michael. She throws one ball straight up and one ball straight down, but both wit
antoniya [11.8K]

Answer:

Both balls have the same speed.

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4 0
3 years ago
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C.) In this, number of Hydrogen atoms is 4
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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
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