In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.
The matter in gaseous state can be expanded to occupy the volumes of the container.
<h3>
Volume of each of the container</h3>
The volume of each of the container is calculated as follows;
<h3>Volume of the rectangular container</h3>
V = 5 in x 6 in x 3 in
V = 90 in³
<h3>Volume of the cylindrical container</h3>
V = πr²h
V = (π)(2.5 in)²(8 in)
V = 157.1 in³
<h3>Volume of the matter</h3>
Vm = 3 in x 4 in x 5 in
Vm = 60 in³
<h3>Matter in solid and liquid state</h3>
Matter has fixed volume in solid and liquid state.
In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.
<h3>Matter in gaseous state</h3>
Matter has no definite volume in gaseous state.
The matter in gaseous state can be expanded to occupy the volumes of the container.
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Answer:
3ohms
Explanation:
From Ohm's Law
V = IR
V is that voltage = 3volts
I = current = 1amp
R = resistance in ohms
Putting those values into the above formula.
3volts = 1amp×R
Making R the subject
R = 3/1
R = 3ohms
The resistance of the light bulb is 3ohms.
Responder:
35,2 ohm.
Explicación:
Dado:
La resistencia específica del conductor es,
La longitud del conductor es,
El área de la sección transversal del conductor es,
Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.
Por lo tanto, la resistencia se puede expresar como:
Ahora, conecte los valores dados y resuelva para 'R'. Esto da,
Por lo tanto, la resistencia del conductor es de 35,2 ohm.
Answer: 45 A
Explanation:
Primary only protection 3-phase
I = 3 phase kVA / ( 1.723 * V)
I = 30000 / ( 1.732 * 480 ) = 36.085 A
Table 450.3(B)
Currents of 9A or more column
primary only protection = 125%
Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A
Table 450.3(B) Note 1 does not apply, use next smaller Table 240.6(A)
Next smaller = 45 A
Because one object can have more things in it. I learned it a t school today.