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3 years ago

What is the part of the steam engine that does the work? A. The flywheel B. The cylinder C. The piston D. The turbine

Physics

2 answers:

pav-90 [236]3 years ago

5 0

C. The piston

The steam engine is a heat engine that performs mechanical work using steam. The steam engine uses the force produced by steam pressure to push the piston back and forth.

antoniya [11.8K]3 years ago

5 0

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A rock is dropped off a cliff and falls the first half of the distance to the ground in 2.0 seconds. how long will it take to fa

son4ous [18]

Answer:

The answer is 0.83 seconds.

Explanation:

The formula of free fall is following:

$h=1/2*g*t^2$

Where g=9.8 m/s^2 and t=2 seconds, the rock takes:

$h=1/2*9.8*2^2=19.6$

19.6 meters. This is the half distance of the cliff. The whole distance is 39.2 meters. So it takes:

$39.2=1/2*9.8*t^2\\t^2=8\\t=2.83$

2.83 second to fall down completely. The rock takes the second half of the cliff in 0.83 seconds

8 0

3 years ago

Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc

expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0

3 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0

3 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

2 years ago

A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc

Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0

2 years ago