1. When an object is moving away from us, the light from the object is known as redshift, and when an object is moving towards us, the light from the object is known as blueshift.
2. A wavelength increases in size, and its frequency, and energy decrease.
3. The frequency of a wave increases, and its wavelength decreases.
Redshift is an important term for astronomers. The term can be taken literally. The wavelengths of light are stretched and perceived as shifting toward the red portion of the spectrum. The same thing happens to sound waves when the source moves relative to the observer.
As the wave frequency decreases, the wavelength increases as long as the wave velocity remains constant. If the wave speed stays the same as the frequency decreases, it means that fewer wave peaks or troughs pass through a given point in a given time period. The number of complete wavelengths in a given unit of time is called frequency. Frequency and energy decrease with increasing wavelength.
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Answer:
6.32 m/s
Explanation:
KE = 1/2 m v^2
sqrt (2 * KE /m) = v = 6.32 m/s
The energy (in calories) required to power the heart for one day is 268.32 Cal
<h3>How to convert 1 day to seconds</h3>
1 day = 24 h
24 h = 24 × 60 = 1440 mins
1440 mins = 1440 × 60 = 86400 s
Thus,
1 day = 86400 s
<h3>How to determine the energy (in calories) for 1 day (86400 s)</h3>
From the question given,
1 s = 13 J
Therefore,
86400 s = 86400 × 13
86400 s = 1123200 J
Divide by 4186 to express in calories
86400 s = 1123200 / 4186
86400 s = 268.32 Cal
Thus, the energy needed to power the heart for 1 day is 268.32 Cal
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Complete question
Under normal conditions the human heart converts about 13.0 J of chemical energy per second into 1.30 W of mechanical power as it pumps blood throughout the body. (a) Determine the number of Calories required to power the heart for one day given that 1 Calorie equals 4186
It is an real machine because if efficiency is below100% it is real machine but if efficiency is 100% it is an ideal machine
A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °
B. The arrow will go over the branch.
<h3>A. How to determine the angle</h3>
- Range (R) = 74 m
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = ?
R = u²Sine(2θ) / g
74 = 33² × Sine (2θ) / 9.8
Cross multiply
74 × 9.8 = 33² × Sine (2θ)
725.2 = 1098 × Sine (2θ)
Divide both sides by 1098
Sine (2θ) = 725.2 / 1098
Sine (2θ) = 0.6605
Take the inverse of sine
2θ = Sine⁻¹ 0.6605
2θ = 41.3
Divide both sides by 2
θ = 41.3 / 2
θ = 20.7 °
<h3>B. How to determine if the arrow will go over or under the branch</h3>
To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = 20.7 °
- Maximum height (H) = ?
H = u²Sine²θ / 2g
H = [33² × (Sine 20.7)²] / (2 ×9.8)
H = 6.94 m
Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).
Therefore, we can conclude that the arrow will go over the branch
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