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romanna [79]
3 years ago
15

Which type of employee would most likely spend the majority of work time in a vehicle? O a Corrections Officer at a jail O a Dis

patcher at a police department O a Paramedic for a fire department O a Bodyguard for a movie star​
Physics
2 answers:
dsp733 years ago
8 0

Answer:

Paramedic

Explanation:

They will be on the move the whole time

VashaNatasha [74]3 years ago
6 0

C. Paramedic for a fire department

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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
Klio2033 [76]

Answer:

a) u=0.05602\ m.s^{-1}

b) u'=0.0169\ m.s^{-1}

Explanation:

Given:

mass of the ball, m=0.4\ kg

horizontal velocity of the ball, u=10\ m.s^{-1}

mass of the person, M=71\ kg

a)

<u>Using the law of  conservation of momentum:</u>

m.v=(M+m).u

0.4\times 10=(71+0.4)\times u

u=0.05602\ m.s^{-1}

b)

Given:

  • rebound velocity of the ball, v'=7\ m.s^{-1}

Using conservation of momentum,

m.v+M.u=m.v'+M.u'

0.4\times10+71\times 0=0.4\times 7+71\times u'

u'=0.0169\ m.s^{-1}

4 0
3 years ago
Read 2 more answers
A photon of wavelength 3.7 nm Compton scatters from an electron at an angle of 90°. What is the modified wavelength? (Enter your
viva [34]

Answer:

3.70242 nm

Explanation:

Using Compton effect formula

Δλ = ( h / mec) ( 1 - cosθ)

where h is planck constant = 6.62607 × 10 ⁻³⁴ m²kg/s

me, mass of an electron = 9.11 × 10⁻³¹ kg

c is the speed of light = 3 × 10⁸ m/s

Δλ = 6.62607 × 10 ⁻³⁴ m²kg/s / (9.11 × 10⁻³¹ kg × 3 × 10⁸ m/s ) ( 1 - cos 90°) =   0.242 × 10 ⁻¹¹ m = 2.42 × 10⁻¹² m  = 0.00242 nm

modified wavelength = 3.7 nm + 0.00242 nm = 3.70242 nm

7 0
4 years ago
Read 2 more answers
Given a force of 50 N and an acceleration of 5 m/s/s, what is the object’s mass?
damaskus [11]
Baseball? ........................sorry if it’s wrong
4 0
3 years ago
Read 2 more answers
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the thro
koban [17]

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

           

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

4 0
3 years ago
When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the alors w
Triss [41]

Answer:

A)  λ = 4.88 10² nm,  B)   λ = 4.08 10² nm

Explanation:

The spectrum of hydrogen is correctly explained by the Bohr model, where the energy of each level is

          Eₙ = -13.606 /n²       [eV]

the transition generally occurs from a given level to a lower state nf <no, so a transition is

          ΔE = E_f -Eₙ = -13,606 ( \frac{1}{n_f^2} - \frac{1}{n_o^2} )

to find the wavelength let's use the planck relation

          ΔE = h f

the speed of light is

          c = λ f

we substitute

          ΔE = h c /λ

          λ = \frac{h \ c}{ \Delta \lambda}

       

let's apply this equation to our case

the Balmer series has as final state the level n_f = 2

A) initial state n₀ = 4, final state n_f = 2

         ΔE = -13.606 ( \frac{1}{2^2} - \frac{1}{4^2} )

         ΔE = 2.55 eV

let's reduce to SI units

         ΔE = 2.55 eV (1.6 10⁻¹⁹ J / 1 eV) = 4.08 10⁻¹⁹ J

     

we calculate

         λ = 6.63 10⁻³⁴ 3 10⁸ / 4.08 10⁻¹⁹

         λ = 4.875 10⁻⁻⁷ m

we reduce to nm

         λ = 4.875 10⁻⁷ m (10⁹ nm / 1m)

         λ = 487.5 nm

we reduce to three significant figures

         λ = 4.88 10² nm

B) initial state n₀ = 5

          ΔE = -13,606 ( \frac{1}{2^2}  - \frac{1}{5^2} )

          ΔE = 2,857 eV

we repeat the process of the previous point

         ΔE= 2,857 1.6 10⁺¹⁹ = 4.286 10⁻¹⁹J

we look for the wavelength

           λ = 6.63 10⁻³⁴ 3 10⁸ / 4.88 10⁻¹⁹

           λ = 4.0758 10⁻⁷ m

we reduce to nm

           λ = 4.0758 10² nm

ignificant numbers

           λ = 4.08 10² nm

4 0
3 years ago
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