Ion 2 23

A 625 g basketball and a 58.5 g tennis ball are dropped from a height of d = 1.5 m onto the floor. The coefficient of restitutio

stich3 [128]

Answer:

Maximum height=7.3535 m

Explanation:

Solution of the problem is given in the attachments.

3 0

3 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago

Two flat plates, separated by a space of 4 mm, are moving relative to each other at a velocity of 5 m/sec. The space between the

xenn [34]

Answer:

0.008

Explanation:

From the question, the parameters given are:

Velocity V = 5 m/s

Pressure = 10 pa

But pressure = F/A

10 = F/A

F = 10A

Substitute all the parameters into the formula below

Coefficient of viscosity (η) = F × r /[AV]

Where

F = tangential force,

r = distance between layers,

A = Area, and

V = velocity

(η) = 10A × 0.004 /[A × 5]

The A will cancel out

(η) = 10 × 0.004 /[5]

(η) = 0.04 /5

(η) = 0.008

Therefore, the coefficient of viscosity of the fluid is 0.008

5 0

2 years ago

Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes

Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

At the inlet of turbine P=10 MPa ,T=500 C

AT the exit of turbine P=10 KPa ,x=0.9

Required power=5 MW

From steam table

h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)

$h_g$ =2675.1 KJ/Kg, $h_f$ =417.51 KJ/Kg

$s_g$ = 7.3 KJ/Kg-K , $s_f$ =1.3 KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= $h_f+x(h_g- h_f)$

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34 KJ/Kg

Lets take m is the mass flow rate of steam

So $5\times 10^3=m\times (3374-2449.34)$

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0

3 years ago

Which of the following fluid power systems is portable and human-scale?

meriva

Answer:

a fluid power engine okk done please

7 0

2 years ago

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