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Gala2k [10]
3 years ago
5

A motor vehicle has a mass of 1200kg and the road wheels have a radius of 360mm. The engine rotating parts have a moment of iner

tia of 0.36kgm2 and the four wheels together have a moment of inertia of 4.8kgm2. The gear ratio from engine to back axle is 6 and when the vehicle speed is v m/s, the engine torque available for propulsion is 82Nm and the resistance to motion is (240 +0?)N. Determine the acceleration of the vehicle at a speed of 15m/s on a level road and the time required to increase the speed of the vehicle from 15 to 25m/s​
Engineering
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

manda a senha do brainly bloquearam os amigos e você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo uma academia de lima e você vai ver o meu perfil completo a minha amiga Jeciane estamos completo a minha amiga Jeciane estamos fazendo uma academia de lima e você vai ver o meu perfil você vai ver o meu perfil você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e fazendo uma academia de lima e você vai ver o meu perfil completo a minha amiga Jeciane estamos fazendo

Explanation:

Marcar como melhor porfavo

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The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought
harkovskaia [24]

Answer:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, M_s = 600 g

mass of dry soil sample, M_d = 200 g

mass of water content, M_w = M_s - M_d = 600 g - 200 g = 400 g

The water content is determined as;

M_w(\%)  = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%)  = \frac{600-200}{600} *100 \% \\\\M_w(\%)  = 66.67 \%

Therefore, the water of the saturated clayed soil is 66.67 %.

6 0
3 years ago
Demonstreaza in 20 de propoziti ca snoava pacala si zarzarele boerului e o snoava
S_A_V [24]

Answer:

oops i dnt understand this language.

4 0
3 years ago
A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h i
dmitriy555 [2]

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

{v}^{2}  -  {u}^{2}  = 2as \\  {20}^{2}  -  {10}^{2}  = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a =  \frac{300}{200 }  =  \frac{3}{2}  \: m {s}^{ - 2}

Now, we substitute v, u, and a in the formula

v = u + at \\ 20 = 10 +  \frac{3}{2} t \\  \frac{3}{2} t = 10 \\ 3t = 20 \\ t =  \frac{20}{3}  = 6.67 \: seconds

Please mark Brainliest if this helps!

6 0
3 years ago
How far do you jog each morning? You prefer to jog in different locations each day and do not have a pedometer to measure your d
Mashutka [201]

Answer:

The program is given below with appropriate comments for better understanding

Explanation:

#Program

# foot stride = 2.5 feet

# 1 mile = 5280 feet

no_stride_first_min = int(input('Enter the number strides made durng the first minute of jogging: '))

no_stride_last_min = int(input('Enter the number strides made durng the last minute of jogging: '))

avg_stride_one_min = (no_stride_first_min + no_stride_last_min)/2 # calculates the average stride per minute

jogging_duration = float(input('Enter the total time spent jogging in hours and minute: '))

jogging_duration_hours = int(jogging_duration) # gets the hour

jogging_duration_min = jogging_duration - int(jogging_duration) # gets the minute

tot_jogging_duration_min = jogging_duration_hours*60 + jogging_duration_min # calculates total time in minutes

dist_feet = (avg_stride_one_min*2.5)*tot_jogging_duration_min # calculates the total distance in feet

dist_miles = dist_feet/5280 # calculates the total distance in mile

print('Distance traveled in miles = {0:.2f} miles'.format(dist_miles))

7 0
3 years ago
3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the
gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

                          V_fat = x_fat*V

- The volume of the muscle is given by:

                          V_muscle = V - V_fat

                                            = V - x_fat*V

                                            =V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

                         mass_fat + mass_muscle = Total average mass

                         p_fat*V_fat + p_muscle*V_muscle = p_avg*V

                         p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

                         Weight reading (Wsa) = m = p_1*V

                         p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

                         Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

                         Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

                         Where, p_2 = p_1 - p_water

                         p_2 = Wsw / V

- The average density p_avg:

                         p_avg = 0.5*(p_1 + p_2)  

                         p_avg = 0.5*(Wsa / V + Wsw / V)  

                         p_avg = 0.5*(Wsa + Wsw) / V                      

- Plug in the mass equation:

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

                         x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

                   x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

                         

6 0
3 years ago
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