All categories

3 years ago

Which option demonstrates when most vehicles lose their efficiency?

Engineering

2 answers:

faltersainse [42]3 years ago

7 0

Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

guapka [62]3 years ago

6 0

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

If u think this helped u ..plz mark me as brainliest ..

And follow me

You might be interested in

Multiple Choice

mote1985 [20]

I think it’s manufacturing

7 0

3 years ago

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool

MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)

Q = 95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C

5 0

3 years ago

Two particles have a mass of 7.8 kg and 11.4 kg , respectively. A. If they are 800 mm apart, determine the force of gravity acti

aleksley [76]

Answer:

A) About $9.273 \times 10^{-9}$ newtons

B) 76.518 newtons

C) 111.834 newtons

Explanation:

A) $F_g=\dfrac{GM_1M_2}{r^2}$ , where G is the universal gravitational constant, M 1 and 2 are the masses of both objects in kilograms, and r is the radius in meters. Plugging in the given numbers, you get:

$F_g=\dfrac{(6.67408 \times 10^{-11})(7.8)(11.4)}{(0.8)^2}\approx 9.273 \times 10^{-9}$

B) You can find the weight of each object on Earth because you know the approximate acceleration due to gravity is 9.81m/s^2. Multiplying this by the mass of each object, you get a weight for the first particle of 76.518 newtons.

C) You can do a similar thing to the previous particle and find that its weight is 11.4*9.81=111.834 newtons.

Hope this helps!

3 0

3 years ago

Which statements describe the motion of car A and car B? Check all that apply. Car A and car B are both moving toward the origin

vekshin1

Answer:

car a is moving faster than the car b

8 0

3 years ago

Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30

shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0

3 years ago