ENGINEERING COLLEGE

Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car 

Answers

Answer 1
Answer:

Answer:

3 m/s²

Explanation:

Acceleration is calculated as :

a= Δv/ t

where ;

Δv = change in velocity

Δv = 45 - 0 = 45  m/s

t= 15 s

a= 45 /15

a= 3 m/s²

Related Questions

At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?
The answer to the question mark the park in a
You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel
The 50mm diameter cylinder is made from Am 1004-T61 magnesium (E = 44.7GPa, a = 26x10^-6/°C)and is placed in the clamp when the temperature is T1 = 15°C. If the two 304-stainless-steel (E =193GPa, a = 17x10^-6/°C) carriage bolts of the clamp each have a diameter of 10mm, and they holdthe cylinder snug with a negligible force against the rigid jaws, determine the temperature at whichthe average normal stress in either the magnesium or steel becomes 12 MPa.
Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output

A 78-percent efficient 12-hp pump is pumping water from a lake to a nearby pool at a rate of 1.5 ft3/s through a constant-diameter pipe. The free surface of the pool is 32 ft above that of the lake. Determine the irreversible head loss of the piping system, in ft, and the mechanical power used to overcome it. Take the density of water to be 62.4 lbm/ft3.

Answers

Answer:

irreversible head loss is 38.51 ft

mechanical power 6.55 hp

Explanation:

Given data:

Pump Power 12 hp = 8948.39 watt = 6600 lbs ft/sec

Q = 1.5 ft^3/s

Pump actually delivers P'  = \eta P = 0.78 * 8948.39 = 6979.74 watt

Power that water gains= mgh = \rho \phi gh = r \phi h

P = 62.4 * 1.5 * 32 = 2995.2 lbs ft/sec

hence Power lost = 6600 - 2995.2  3604.8 lbs ft/sec = 6.55 hp

head losshl = (3604.8)/(r \phi) = (3604.8)/(62.4 * 1.5)

hl = 38.51 ft

Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

Answers

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30  + ( bicarbonate ion ) * 50/61

                                   = 15 * (50/30) + 120*( 50/61 )  = 123.361 mg/l  as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57  = 2.692*10^-5  moles/l

[ OH^- ]   = 2.692*10^-5  * 179/mole * 10^3 mg/g  = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3  = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

The pressure in an automobile tire depends on thetemperature of the air in the tire. When the air temperature is25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire whenthe air temperature in the tire rises to 50°C. Also, determinethe amount of air that must be bled off to restore pressure toits original value at this temperature. Assume the atmosphericpressure to be 100 kPa.

Answers

Answer:

The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.

The amount of air that must be bled off to restore pressure 0.007 Kg

Explanation:

Knowing

T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025 m^(3)

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

P1 V1 = m1 R1 T1

m1 = (P1 V1)/(R1 T1) = (310 * 10^(3) * 0.25 )/(287 - 298) = 0.091 Kg\n \n

After the temperature rise

P2 = (m2 * R * T2)/(V2) = (0.091 *287*323 )/(0.025) = 337.43 KPa

after bleeding the pressure and the volume returns  to its first value

P1 = P2 and V1 = V2

then

(m2 * R * T2)/(V) = (m1 * R * T1)/(V)

m2 = (m1*T1)/(T2)

m2 = (0.091*298)/(332) = 0.084 Kg\n\n

mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg

P2 = 337.43 KPa

mbleed = 0.007 Kg

An interest buydown program offers to reduce interest rates by 4% from the base rate. Suppose the base rate for a loan of $8000 is 8% for 10 years. What is the monthly payment before and after the buydown? In this case, use monthly compounding, that is, the term is 120 payment periods, and the interest per month is 0.667% before and 0.333% after the buydown.

Answers

Answer: The monthly payment before the buydown is $71.3

The monthly payment after the buydown is $68.9

Explanation: The payment is compounding so we use compound interest;

A= P[1+(r/n)^nt]

Where;

A= Compounded amount

P = principal

r= interest rate per payment

n= number of payment per period

t= number of period.

NOTE: from our questions, the period is yearly and the payment is monthly. Therefore;

number of payment per period (n) is 12

number of payment period (t) is 10

P=$8000, r= 0.667% or 0.333%

FIND MONTHLY PAYMENT BEFORE BUYDOWN:

Step 1: find the Compounded amount to pay.

A= $8000[1+(0.00667÷12)^(12×10)]=

$8551.64 this is the total amount he has to pay for a period of 10years

Step 2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months

$8551.64÷120= $71.3 monthly

FIND MONTHLY PAYMENT AFTER BUYDOWN:

Step 1: find the compounded amount to pay.

A= 8000[1+(0.00333÷12)^(12×10)=

$8270.85 this is the total amount he has to pay for a period of 10years

Step2: How much does he has to pay monthly for a period of 10year;

Therefore his payment will be for 120 months;

$8270.85÷120= $68.9 monthly

Which of these factors would help the environment?a. Betty drinks only bottled water instead of soda.
b. Greg changes all of the light bulbs in the office to energy efficient bulbs.
c. John keeps his off topic conversations down to 10 minutes a day.
d. Doug turns the air conditioning down to cool the office during the hot summer months.
Please select the best alswer from the choices provided
A
B
D

Answers

Answer:

B. Greg changes all of the light bulbs in the office to energy efficient bulbs.

Explanation:

Energy efficient bulbs help you to reduce the carbon footprint of your office / house and last up to 12 times as long as traditional bulbs, using less electricity to emit the same amount of light as a traditional bulb.

Answer:

b

Explanation:

For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.

Answers

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 =  0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 =  0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97  = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.
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Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L