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Jobisdone [24]
3 years ago
13

A horizontal force of 50 N is required to push a wagon across a sidewalk at a constant speed.

Physics
1 answer:
Lerok [7]3 years ago
5 0
The answer is In fact A because it’s unbalanced
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A gymnast does a one-arm handstand. The humerus, which is the upper arm bone between the elbow and the shoulder joint, may be ap
igor_vitrenko [27]

Answer:

                  = 5.241 \times 10^{-5} m

Explanation:

Given:

 Length of cylinder is, L = 0.27 m

Outer radius of cylinder is, r_out = 1.12×10^{-2} m

Inner radius of cylinder is, r_in = 3.9×10^{-3} m

Mass of person, m = 60 kg

 Young's modulus , Y = 9.4×10^9 N/m2

(a)

     Compressional strain of humerous is,

Strain = \frac{Stress}{Young's\ modulus}

     \frac{\Delta L}{L_0}   = \frac{\frac{F}{A}}{Y}

                  = \frac{(mg)}{\pi(r_out^2 - r_in^2  )Y}

                  = \frac{(60 kg)(9.8 m/s2 )}{(\pi)[( 1.12\times 10^{-2})^2 - (3.9\times 10^{-3} m)^2] (9.4\times 10^9 N/m2 )}[tex]                   [tex]= 1.80\times 10^{-4} m

(b)  Let assume that humerous is compressed by ΔL

       Since,   strain = ΔL/L0

      (1.80 \times 10^{-4} m) = ΔL / 0.29 m

     Thus,

           ΔL = (4.56 \times 10^{-4} m)(0.29 m)

                  = 5.241 \times 10^{-5} m

7 0
3 years ago
A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug
vlada-n [284]

Answer:

The Minimum wavelength is  \lambda_{min}= 382.2nm

The Maximum wavelength is \lambda_{max}= 624.2nm

Explanation:

From the question we are told that  

              The energy range is  E_r = 3.25eV \ and  \ 19.9eV

   Considering E = 19.9eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 19.9 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{max}}

Here h is the Planck's constant with value of  h= 6.625*10^{-34}J\cdot s

        c is the speed of light with value of  c = 3*10^8 m/s

Substituting values and making \lambda the subject of the formula

                       \lambda_{max} = \frac{hc}{E}

                         = \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}

                         \lambda_{max}= 624.2nm

  Considering E = 3.25eV

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

              E = 19.9eV = 3.25 *1.6 *10^{-19}J

This energy is mathematically represented as

                    E = \frac{hc}{\lambda_{min}}

Substituting values and making \lambda the subject of the formula

                       \lambda_{min} = \frac{hc}{E}

                           = \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}

                           \lambda_{min}= 382.2nm

3 0
3 years ago
In one cycle, a freezer uses 585 J of electrical energy in order to remove 1750 J of heat from its freezer compartment at 10.0 F
gtnhenbr [62]

Answer:

(a) 0.525

(b) 1750J

Explanation:

Coefficient of Performance (COP) = Q2/(Q1 - Q2)

Q1 = heat rejected to the surrounding = 1750J

Q2 = heat absorbed in the environment = 585K

COP = 585/(1700 - 585) = 585/1115 = 0.525

(b). Heat expelled to the surrounding (room) = 1750J

7 0
3 years ago
Scientists usually write reports about their investigations. Why do they do this?
worty [1.4K]
As for me, the reson why scientists usually write reports about their investigations is because they want to <span>d. to communicate their work to other scientist. As far as I am concerned in this sphere they have to get many other opnions besides their own in order to complete or expand one that they have made up. Hope it helps :)</span>
5 0
3 years ago
If the wind or current is pushing your boat away from the dock as you prepare to dock, which line should you secure first?
Akimi4 [234]

Answer:

Bow Line

Explanation:

If the wind or current is pushing your boat away from the dock, bow line should be secured first.

1- We should cast off the bow and stern lines.

2-With the help of an oar or boat hook, keep the boat clear of the dock.

3-Leave the boat on its own for sometime and let the wind or current carry the boat away from the dock.

4 - As you see there is sufficient clearance, shift into forward gear and slowly leave the area.

7 0
3 years ago
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