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Pavlova-9 [17]
3 years ago
11

A(n) 82.7 kg boxer has his first match in the Canal Zone with gravitational acceleration 9.782 m/s 2 and his second match at the

North Pole with gravitational acceleration 9.832 m/s 2 . a) What is his mass in the Canal Zone? Answer in units of kg.
Physics
1 answer:
Annette [7]3 years ago
6 0

Answer:

82.7 kg

Explanation:

the mass of the boxer remains unchanged, this is because mass is a measure of the quantity of matter in an object irrespective of its location and the gravitational force acting at its location. this means mass is independent of the gravitational acceleration hence it remains the same 82.7 kg. its unit is in kilograms (Kg).

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Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstro
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Answer:

Explanation:

The formula for hydrogen atomic  spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 ,   ...   etc

energy of first line

E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

7 0
3 years ago
26. The speed of light in a certain medium is
horrorfan [7]

The number we need in order to answer the question belongs in the space between the words "is" and "of".  You left that blank blank, so there really isn't any question here to answer.

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How long would this acceleration last
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It would last as long as the applied force continued, or until the accelerating object hit something.
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during a workout, a football player pushes a blocking dummy a distance of 30 m. while pushing the dummy the same distance a seco
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Power=\frac{Work}{Time}=\frac{Force\times distance}{time}

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8 0
3 years ago
Read 2 more answers
A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight
Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is  7.61 x 10⁻⁶  

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

  • m is mass supported by the steel
  • g is acceleration due to gravity
  • A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

  • E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

  • ΔL is change in length
  • L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0
1 year ago
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