1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kotykmax [81]
3 years ago
5

An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50

km/h? If the vehicle accelerates to 100 km/h in 15 seconds, what is the change in power required?
Engineering
1 answer:
netineya [11]3 years ago
3 0

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

Ek=0.5mv^{2}

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

Ek=0.5(1200)(13.89)^2

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

Ek=0.5(1200)(27.78)^2=462963J\\

taking into account all of the above the following equation is inferred

ΔE=\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15}  =23146.916W=23.15kW

You might be interested in
The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to
skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 kN/m^{2}

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = 0.02\ m^{3}/s

(Since, 1 litre = 10^{-3} m^{3})

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, H_{loss} = 45 cm = 0.45\ m

Height, h_{roof} = 2.5\ m

Now,

The velocity in the bore is given by:

v = \frac{Q}{\pi (\frac{d}{2})^{2}}

v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s

Now, using Bernoulli's eqn:

\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k                  (1)

The velocity head is given by:

\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553

Now, by using energy conservation on the surface of water on the roof and that in the tank :

\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}

\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45

P_{tank} = 5.5\times \rho \times g

P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}

4 0
3 years ago
PLSSSSS Help !!!!!!!!!!!. It's due today. <br> I will give brainliest to correct answer
4vir4ik [10]

Answer:

b

Explanation:

3 0
3 years ago
Read 2 more answers
A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

3 0
3 years ago
Lithium at 20°C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in na
Nutka1998 [239]

Answer:

the atomic radius of a lithium atom is 0.152 nm

Explanation:

Given data in question

structure = BCC

lattice constant  (a) = 0.35092 nm

to find out

atomic radius of a lithium atom

solution

we know structure is BCC

for BCC radius formula is \sqrt{3} /4 × a

here we have known a value so we put a in radius formula

radius =  \sqrt{3} /4 × a

radius =  \sqrt{3} /4 × 0.35092

radius = 0.152 nm

so the atomic radius of a lithium atom is 0.152 nm

5 0
3 years ago
Which of the following technologies might affect how a vehicle is driven
yulyashka [42]

Explanation:

bhummmm like this message

8 0
3 years ago
Other questions:
  • Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This
    7·1 answer
  • Write cout statements with stream manipulators that perform the following:
    8·1 answer
  • The viscosity of the water was 2.3×10^−5lb⋅⋅s/ft^2 and the water density was 1.94 slugs/ft^3. Estimate the drag on an 88-ft diam
    13·1 answer
  • If the resistance reading on a DMM'S meter face is to 22.5 ohms in the range selector switch is set to R X 100 range, what is th
    5·1 answer
  • Barry wants to convert mechanical energy into electric energy. What can he use?
    5·2 answers
  • Which of the following power generation technologies produces a direct current?
    9·1 answer
  • Mining is an example of this type of business
    7·1 answer
  • an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required
    6·2 answers
  • An engineer testifying as an expert witness in a product liability case should:
    11·1 answer
  • How many gallons of 25% concentrated due will Lucy need to add to the 15% concentrated dye to make a batch with a concentration
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!