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ankoles [38]
3 years ago
14

When starting a car some energy is used to increase the fuels temperature during ignition.this energy causing combustion is a:ab

sorbed energyb:chemical energy c:activation energy d:latent energy
Physics
1 answer:
s2008m [1.1K]3 years ago
5 0
Internal combustion engine works by heat. The chemical energy of the fuel turns into heat energy when the fuel is burned, which produces mechanical energy to drive the pistons, to manage the column of the facility and run the vehicle on the road.
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What does the evidence in this passage suggest?
mestny [16]

Answer: D

Explanation:

cause i got it right

4 0
2 years ago
Which of the following is a physical property?
coldgirl [10]
Flammability, I think!
6 0
3 years ago
Read 2 more answers
A 3047.8 kg truck has lost its brakes coming down a mountain. Fortunately, there is a ramp of thick gravel inclined at 9.5 degre
Yuliya22 [10]

Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

<u>Step 3: </u>Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

<u>Step 4:</u>  What work is done by the truck on the gravel?  

Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ

5 0
3 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0
3 years ago
(HELP!! 10pts) a train travels 75km in 1hr, and then 68km in 2hrs. what is its average speed?
blondinia [14]
Average speed = total distance travelled ÷ total time taken

AS = (75km + 68km) ÷ (1hr + 2hr)

As = 143km ÷ 3hr

AS = 47.66667 km/hr

AS = 47.7 km/hr (3sf)
8 0
3 years ago
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