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2 years ago

Will mark brainlist

Physics

1 answer:

Ksenya-84 [330]2 years ago

4 0

Answer:

5.00 x 10^9 j

Explanation:

mass = 100 kg

speed / velocity = 10 km/s x 1000 = 10,000m

K.E =

$\frac{m{v}^{2} }{2}$

10,000x 100/ 2

= 5000000000

= 5.0 x 10^9j

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For a molecule of O2 at room temperature (300 K), calculate the average angular velocity for rotations about the x or y axes. Th

Ksivusya [100]

Answer: 6.65 rad/s

Explanation:

Firstly, we need to know that according the kinetic theory of gases, the average kinetic energy $KE$ of a molecule with $i$ degrees of freedom is:

$KE=\frac{i}{2}kT$ (1)

Where:

$i=5$ because oxigen is a diatomic molecule and has 5 degrees of freedom

$k=1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K}$ is the Boltzman constant

$T=300 K$ is the temperature

Then:

$KE=\frac{5}{2}(1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K})(300 K)$ (2)

$KE=1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}$ (3) With this value of average kinetic energy we can find the average angular velocity, with the following equation:

$KE=\frac{1}{2}I \omega^{2}$ (4)

Where:

$I$ is the moment of inertia

$\omega$ is the angular velocity

Now, $I=m R^{2}$ (5)

Being $m=31.98 g/mol=0.03198 kg/mol$ the molar mass of Oxigen molecule and $R=0.121(10)^{-9}m$ the distance between atoms

$I=(0.03198 kg/mol)(0.121(10)^{-9}m)^{2}$ (6)

$I=4.682(10)^{-22} \frac{kg}{mol} m^{2}$ (7)

Substituting (7) and (3) in (4):

$1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}=\frac{1}{2} (4.682(10)^{-22} \frac{kg}{mol} m^{2}) \omega^{2}$ (8)

Finding $\omega$ :

$\omega=6.65 rad/s$ (9) This is the average angular velocity for a molecule of O2

7 0

3 years ago

A 6 kg bowling ball moves with a speed of 3 m/s. How fast does a 7 kg bowling ball need to move so that it has the same kinetic

maw [93]

Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion.

$K.E=\frac{1}{2}mv^2$

m = mass of object

v= velocity of the object

$K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules$

b) for a 7 kg bowl to have kinetic energy of 27 Joules:

$27J=\frac{1}{2}\times 7kg\times v^2$

$v^2=7.7$

$v=2.8m/s$

Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy

4 0

2 years ago

A football is thrown with an acceleration of 15 m/s^2 and a force of 13 N. What is its mass?

-Dominant- [34]

I believe it is b. Lmk if I’m wrong

7 0

2 years ago

Read 2 more answers

What is the net force required to accelerate a 16 kg box at a rate or 1.4 m/s^2

stich3 [128]

You can use Newton's Second Law which states:
$\sum \vec{F} = m\vec{a}$

Plug in given information:
$\sum \vec{F}=16(1.4) = 22.4N$

This is closest to option b which is your answer.

8 0

2 years ago

Un objeto de 200 gramos está amarrado del extremo de una cuerda y gira describiendo un círculo horizontal de 1.20 m de radio a r

vesna_86 [32]

Answer:

La tensión es 85.3 N.

Explanation:

Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:

$\Sigma F_{x} =ma_{c}$

$T = ma_{c}$

Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.

En donde:

m: es la masa del objeto = 200 g = 0.200 kg

$a_{c}$ : es la aceleración centrípeta

La aceleración centrípeta viene dada por:

$a_{c} = \omega^{2} r$

En donde:

ω: es la velocidad angular del objeto = 3 rev/s

r: es el radio = 1.20 m

Entonces, la tensión es:

$T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N$

Por lo tanto, la tensión es 85.3 N.

Espero que te sea de utilidad!

5 0

2 years ago