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Margaret [11]
3 years ago
11

B all figures represents molecules

Physics
1 answer:
Volgvan3 years ago
5 0
C. Figure 11 and figure 1v are both compounds
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A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta
nevsk [136]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

4 0
3 years ago
In fig. 30.11, suppose that and with switch open, switch is left closed until a constant current is established. then is closed
Gnom [1K]
In a particle accelerator a positron (C= +1.6 x 10-19) travels through a perpendicular magnet field with a magnitude of 3.1 x 10-2 T. At what speed must the positron travel in order for it to experience a force of 4.75 x 10-14 N? 28. An alpha particle (2 protons and 2 neutrons) experiences a downward force of 2.9 x 10-14 N while traveling in a magnetic field with a strength of 5.1 x 10-19 T pointing to the north. Find the speed of the particle and the direction it must be traveling in. 29. Find the length of a wire if it experiences a .63N force when it travels through a magnetic field with a strength of 0.85T whilst carrying 5.0 amps of current. 30. A coil with 462 turns of wire, a total resistance of 36Ω , and a cross-sectional area of 0.25 m2 is positioned with its plane perpendicular to the field of a powerful electromagnet. What average current is induced in the coil during the 0.37s that the magnetic field drops from 3.1 T to 0.0 T? 31. A step-up transformer has a potential difference across the primary of 28 V and a potential difference across the secondary of 3.0 × 104 V. There are 28 turns in the primary coil. How many turns are in the secondary? 32. A step-up transformer is used to create a potential difference of 1.6872 × 105 V across the secondary. The potentia
3 0
3 years ago
In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t
Tcecarenko [31]

Answer:

The amount of heat transfer is 21,000J .

Explanation:

The equation form of thermodynamics is,

ΔQ=ΔU+W

Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.

Substitute 0 J for W and 0 J for ΔU

ΔQ = 0J+0J

ΔQ = 0J

The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change

The heat transfer is,

ΔQ=Q  (in ) −Q (out )

Substitute 19000 J + 2000 J for Q(in)  and 0 J for Q(out)

ΔQ=(19000J+2000J)−(0J)

=21,000J

​Thus, the amount of heat transfer is 21,000J .

​

8 0
3 years ago
Read 2 more answers
Si tenemos tres cubos del mismo tamaño (hierro, madera e icopor). ¿Qué diferencias puede encontrar entre ellos?
cricket20 [7]

Answer: La diferencia es el peso (o la masa), siendo que el cubo de hierro es el mas pesado, después viene el de madera y después el de icopor.

Explanation:

Ok, los 3 cubos tienen el mismo tamaño, lo que implica que tienen el mismo volumen.

Ahora es útil recordar la relación:

Densidad = masa/volumen.

Masa = densidad*volumen.

Nosotros sabemos que la densidad del hierro es mas grande que la de la madera, y la densidad de la madera es mas grande que la del icopor.

Entonces, por la relación anterior, y sabiendo que todos los cubos tienen el mismo volumen, podemos reconocer que el cubo de hierro tiene mayor masa, después viene el de madera y después viene el de icopor.

Y sabiendo que:

masa*gravedad = peso

podemos saber que el cubo mas pesado es el de hierro, después el de madera y después el de icopor.

Además de esta diferencia, también hay otras que no dependen tanto del tamaño del objeto, como pueden ser las capacidades caloríficas, el como reaccionan a campos eléctricos y cosas así que son triviales, pues son diferentes para casi todos los materiales.

5 0
3 years ago
The equation that is used to solve second law problems is # F= ma.
maw [93]
F= Force 
M=Mass
A= acceleration 
F=N 
Mass= in grams or kilo grams (mostly kg)
A= m/s 

8 0
3 years ago
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