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nikklg [1K]
2 years ago
13

Pls help me Will mark you as brain plssss help me for these 2 questions just say top or bottom for which answers you pick

Physics
1 answer:
DaniilM [7]2 years ago
8 0

Answer:

Top(Question 4): A. Increase the duration of walking each day

Bottom(Question 5): C. Evaluate and adjust her choice of activities, frequency, duration, and intensity

Explanation:

<u>Question 4: </u>The principle of progression states that as your body adapts to your exercise routine, you have to change it up. This can mean gradually increasing the weight, duration, or intensity of your weight training to see growth.

Hope this helps!! <3

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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
Sedbober [7]

Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

3 0
3 years ago
How does surfing relate to physics
uranmaximum [27]
It’s physical something you can physically do
5 0
3 years ago
Based on your observations of the six collisions, describe the physical difference between elastic and inelastic collisions.
Ahat [919]

Answer:

Collisions are basically two types: Elastic, and inelastic collision. Elastic collision is defined as the colliding objects return quickly without undergoing any heat generation. Inelastic collision is defined as the where heat is generated, and colliding objects are distorted.

In elastic collision, the total kinetic energy, momentum are conserved, and there is no wasting of energy occurs. Swinging balls is the good example of elastic collision. In inelastic collision, the energy is not conserved it changes from one form to another for example thermal energy or sound energy. Automobile collision is good example, of inelastic collision.

5 0
2 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
A person is just as likely to become dehydrated in the cold as in the heat. Please select the best answer from the choices provi
alexdok [17]
The answer would be A) TRUE because we as humans can be dehydrated easily just by working out in sun or working out in the cold winter, that's how our body burns fat and calories, but it also burns our eyes since sweat has salt in it, and dehydrated means to run out of water for us, like not enough water, but yes to the answer its T.
8 0
3 years ago
Read 2 more answers
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