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Mashcka [7]
3 years ago
12

4. What is the speed of a walking person in m/s if the person travels 1000m in 20 minutes?

Physics
1 answer:
kramer3 years ago
7 0

Answer:

0.80 m/s

Explanation:

What is the speed of a walking person in m/s if the person travels 1000 m in 20 minutes?

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Help me I do not understand
ololo11 [35]

Answer:

170N

Explanation:

First add 530N to 150N and you get 680N, then add 400N to 450N and get 850N. So subtract 850N by 680N and you get 170N

4 0
4 years ago
A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou
prisoha [69]

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

tan\theta =\frac{v^{2}}{rg}

tan\theta =\frac{18.61^{2}}{53.1\times 9.8}

tan  θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

6 0
3 years ago
An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The val
OverLord2011 [107]

Answer:

Answer

The Final Quality of teh R-134a in the container  is  0.5056

The Total Heat transfer is Q_{in} = 22.62 KJ

Explanation:

Explanation is  in the following  attachments

3 0
3 years ago
You are playing a violin, where the fundamental frequency of one of the strings is 440 Hz, as you are standing in front of the o
Natalka [10]

Answer:

a)   L = 440 cm

Explanation:

In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are

             λ₁ = 4L             fundamental

             λ₃ = 4L / 3       third harmonic

             λ₅ = 4L / 5       five harmonic

             

The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are

              λ₁ = 2L                    fundamental

             λ₂ = 2L / 2              second harmonic

             λ₃ = 2L / 3              third harmonic

             λ₄= 2L / 4               fourth harmonic

They indicate that resonance occurs in the fourth harmonic, let's look for the frequency

              v =λ f

for the fundamental

              v = λ₀ f₀

              V = 2L f₀

for the fourth harmonica

              v = λ₄  f ’

              v = L / 2  f'

             2L f₀ = L / 2 f ’

             f ’= 4 f₀

             f ’= 4 440

             f ’= 1760 Hz

for this frequency it has the resonance with the tube

           f ’= 4L

           L = f ’/ 4

           L = 1760/4

           L = 440 cm

b) let's find the frequency of the next harmonic in the tube

             λ₃ = 4L / 3

             λ₃ = 4 400/3

             λ₃ = 586.6 cm

            v = λf

            f = v / λlam₃

            f₃3 = 340 / 586.6

            f3 = 0.579

as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances

6 0
4 years ago
What is the name given to the specific latent heat of a
SCORPION-xisa [38]

Answer:

the heat of fusion

Explanation:

7 0
3 years ago
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