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3 years ago

Match the following.

Physics

1 answer:

anastassius [24]3 years ago

8 0

Answer:

You should really put the definitions man

Explanation:

The question is Match the following.

1. intake stroke

a.the gas vapor/air mixture enters the cylinder

2. compression stroke

b. the gas vapor/air mixture is pressurized

3. power stroke

c. hot gases are expelled from the cylinder

4. exhaust stroke

d. a spark from a spark plug initiates a chemical reaction

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In a binary-star system that produces a nova, the white dwarf pulls matter from the companion star. The matter forms an accretio

Studentka2010 [4]

As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.

Now the process:

First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.

After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.

As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.

4 0

2 years ago

Gravity on Jupiter is 25 m/s/s, what is the weight of a 12 kg object on Jupiter?

stiv31 [10]

Answer:

300 Newtons

Explanation:

Weight is the force of attraction between two bodies, one usually larger (like a planet), and one smaller (like a person). Force can be calculated using the formula: Force = mass × acceleration.

The mass here is 12kg, the acceleration, which in this case, is the acceleration due to gravity is 25m/s/s, by plugging in our values, we have

Force = 12 × 25 = 300 Newtons or 300 N for short.

3 0

2 years ago

A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is

notka56 [123]

Answer:

a) Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s

Explanation:

a) We have equation of motion s = ut + 0.5at²

Here s = 400 radians

t = 8 s

a = 5 rad/s²

Substituting

400 = u x 8 + 0.5 x 5 x 8²

u = 30 rad/s

Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

Here u = 30 rad/s

t = 8 s

a = 5 rad/s²

Substituting

v = 30 + 5 x 8 = 70 rad/s

Final angular speed = 70 rad/s

8 0

3 years ago

a train travles at a speed of 30m/s. the train starts at an initial position of 1000 meters and travels for 30 seconds. what is

ycow [4]

Answer:

1900 meters

Explanation:

30m/s x 30 second = 900 meters

+ 1000 meters starting position

= 1900meters

5 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago