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3 years ago

What kind of engine do most cars use today

Physics

2 answers:

Nadusha1986 [10]3 years ago

4 0

Answer:

4 stroke 4 cylinder Natural Aspirated Gasoline engine

Explanation:

im not 100% sure

VMariaS [17]3 years ago

4 0

Answer:

BMW king of engine do most cars use today

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What conclusions can be drawn about the existence of carbon-12, carbon-13, and carbon-14?

SashulF [63]

<span>These are isotopes of carbon and they all contain 6 protons and 6 electrons but each contains a difference number of neutrons - 6, 7, and 8 respectively.

^ This is the answer because an isotope changes the atomic mass, NOT atomic number. That means that the neutrons are changed, not the protons. </span>

7 0

3 years ago

Read 2 more answers

A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of

igomit [66]

Answer:t=0.81 s

Explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore $F_{net}$ is zero on Penguin

$F_{net}=mg\sin \theta -f_r$

$f_r=$ friction Force

$f_r=\mu mg$

$\mu =$ coefficient of Kinetic friction

$mg\sin \theta =\mu mg$

$\tan \theta =\mu$

$\mu =\tan 9.85=0.45$

after reaching on floor final velocity of penguin will be zero after time t

thus

$v=u+at$

here $a=\mu g$

$a=0.45\times 9.8=4.41$ (deceleration)

$0=3.57-4.41\times t$

$t=\frac{3.57}{4.41}=0.809$

$t=0.81 s$

8 0

3 years ago

At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$