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Tema [17]
3 years ago
15

A car increases its speed at a constant rate of 1.5 m/s2 as it rounds a curve of radius 200 m. The magnitude of the total accele

ration is 2.5 m/s2 . What is the car’s speed at this instant?

Engineering
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

The car's instant speed is 20m/s

Explanation:

a = at + an

The above equation is a vector equation

a is the resultant acceleration

at   is the tangent acceleration

an  is the normal acceleration

See the attached diagram for solution

<h2 />

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Explanation:

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3 years ago
An axial compressive load of 708 kN is applied to a cylindrical component, 81 mm in diameter and 418 mm long, made of aluminium.
dalvyx [7]

Answer:

The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)

Explanation:

The compressive stress in a cylindrical  component can be calculated aby dividing the compressive force F to the cross sectional area A:

fc= F/A

If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.

For acylindrical component the cross sectional area A is:

A=πR^

If the diameter of the component is 81 mm, the radius is the half:

R=81mm /2 = 40.5 mm

Then A result:

A= 3.14 * (40.5 mm)^2  = 5150.4 mm^2

In square meters:

A= 3.14 * (0.0405 m)^2  = 0.005150 m^2

Replacing 708 kN to the force:

fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2

Using the force in Newtons:

F= 70800 N

Finally the compressive stress in Pa is:

fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa

4 0
3 years ago
Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured
Paladinen [302]

Answer:  The normalized corrected value at 32.5 mm = 0.69 mV

Explanation:

Signal value V1 = 20.7 mV at

distance value = 29 mm and

V2 = 15.8 mV is measured at 32.5 mm. 

It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement

V1 = 20.7 - 5.1 = 15.6 mV

V2 = 15.8 - 5.1 = 10.7mV

The normalized corrected value at 32.5 mm will be

Vn = V2/V1 since V1 is the maximum value

Vn = 10.7/15.6 = 0.69 mV

3 0
3 years ago
Compare the direction of the gear wheels in a simple gear train (no idler) with the direction of the gear wheels in a chain driv
Inessa [10]

Answer:

Probably North

Explanation:

6 0
3 years ago
Read 2 more answers
Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 ·K, deter- mine the temperature of the
Aleks [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

a) 95.80°C

b) 8.23 MW

Explanation:

Convection heat transfer coefficient = 860 W/m^2 . k

<u>a) Calculate for the temp of sheet metal when it leaves the oil bath </u>

<em>first step : find the Biot number </em>

Bi = hLc / K  ------- ( 1 )

where : h = 860 W/m^2 , Lc = 0.0025 m ,  K = 60.5 W/m°C

Input values into equation 1 above

Bi = 0.036 which is < 1  ( hence lumped parameter analysis can be applied )

<em>next : find the time constant </em>

t ( time constant ) = h / P*Cp *Lc  --------- ( 2 )

where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C

Input values into equation 2 above

t ( time constant ) = 0.10092 s^-1

<em>Determine the elapsed time </em>

T = L / V = 9/20 = 0.45 min

∴<u>   temp of sheet metal when it leaves the oil bath </u>

= (T(t) - 45 ) / (820 - 45)  = e^-(0.10092 * 27 )

T∞ =  45°C

Ti = 820°C

hence : T(t) = 95.80°C

<u>b) Calculate the required rate of heat removal form the oil </u>

Q = mCp ( Ti - T(t) ) ------------ ( 3 )

m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s

Cp = 434 J/kg°C

Ti =  820°C

T(t) = 95.80°C

Input values into equation 3 above

Q = 8.23 MW

6 0
3 years ago
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