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3 years ago

A positively charged object will attract an object that has

Physics

1 answer:

ser-zykov [4K]3 years ago

5 0

I’d say a negative charge cause opposites attract but I’m not sure

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¿Cuál es la magnitud y dirección de la aceleración de los cuerpos en caída libre en el planeta Tierra?

Tatiana [17]

La magnitud es de 9.8 m/s² ... la aceleración de la gravedad en o cerca de la superficie de la Tierra.

La dirección es hacia el centro de la Tierra. (Llamamos a esa dirección "abajo").

6 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

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3 years ago

A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef

AveGali [126]

The thermal efficiency of an engine is
$\eta= \frac{W}{Q}$
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
$\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%$

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3 years ago

A manager at a local bank analyzed the relationship between monthly salary and three independent variables: length of service (m

Readme [11.4K]

Answer:

Explanation:

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3 years ago

How many kindergarteners do you think you can take on a fight before getting tired or over powered?

yulyashka [42]

Is this a serious question ?

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3 years ago

Read 2 more answers