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Shkiper50 [21]
3 years ago
13

Calculate the power required to accelerate a 1150 kg car from rest to a final velocity of 30 m/s up a 100 m long uphill road wit

h a slope of 30 degrees from horizontal over a time of 12 seconds. Disregard friction, air drag, and rolling resistance.
Engineering
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

power required is 90.131 kW

Explanation:

Given data

mass of car (m) = 1150 kg

starting velocity (v1) = 0 m/s

final velocity (v2) = 30 m/s

slope angle = 30 degrees

slope distance = 100 m

to find out

power required

solution

we know the power =  work / time

here time is given we have calculate the work

work is for car climb 100 m distance is total sum of potential energy + kinetic energy

and we know car climb 100m at 30 degree slope so vertical distance (h) will be h = 100 sin (30) = 50 m

so now potential energy = mass × gravity × height

potential energy = 1150 × 9.81 × 50 = 564075 J

kinetic energy = 1/2 ×mass × velocity²

kinetic energy = 1/2 ×1150 × ( v2 - v1 )²

kinetic energy = 1/2 ×1150 × ( 30 - 0 )² = 517500 J

now put all value in power formula

power = work / time

power = potential energy + kinetic energy / 12

power = 564075  +  517500  / 12 = 90131.25 W

power = 90.131 kW

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ozzi

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Explanation:

8 0
2 years ago
A gas metal arc welder is also known as a _____ welder.<br> A) TIGB) GTAWC) GMAWD) Resistance spot
nignag [31]

Answer:

GMAW

Explanation:

It's literally the initials of that type of welding

7 0
3 years ago
It is important to keeo a copy of your written plan and safety record s off-site. True or false
lyudmila [28]

Answer:

The answer for the question is true

Explanation:

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8 0
3 years ago
At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =
Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

F=ma for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that  W = mg

For simplicity we work with g =9.807 \frac{m}{s^{2}} despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula g = a-bz the "normal" or "standard" weight of an object is given by W = mg = ma when z = 0, so we need to find the value of z that makes W = m(a-bz) = 0.997ma meaning that the original weight decrease by a 0.30%, so now we operate...

m(a-bz) = 0.997ma now we group like terms on the same sides ma(1-0.997) = mbz we cancel equal tems on both sides and obtain that z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m

7 0
3 years ago
Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County a
Vaselesa [24]

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = \frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4}  }  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

5 0
3 years ago
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