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3 years ago

Factor on which quantity of heat depend

Physics

1 answer:

QveST [7]3 years ago

3 0

Answer:

Explanation:

which is the heat capacity per unit mass of a material. Experiments show that the transferred heat depends on three factors: (1) The change in temperature, (2) the mass of the system, and (3) the substance and phase of the substance.

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Is it illegal to attempt gerrymandering

IRINA_888 [86]

Explanation:

The US Supreme Court has affirmed in Miller v. Johnson (1995) that racial gerrymandering is a violation of constitutional rights and upheld decisions against redistricting that is purposely devised based on race. However, the Supreme Court has struggled as to when partisan gerrymandering occurs (Vieth v.

8 0

3 years ago

3.00 m^3 of water is at 20.0°C.

romanna [79]

Answer:

$\triangle V = 0.02484m^3$

Explanation:

Given

$V_1 = 3.00m^3$ --- initial volume

$T_1 = 20.0^oC$ --- initial temperature

$T_2 = 60.0^oC$ --- final temperature

$\gamma = 207*10^{-6$ --- coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

$\triangle V = \gamma * V_2 * (T_2 - T_1)$

So, we have:

$\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)$

$\triangle V = 207 * 10^{-6} * 3.00 * 40.0$

$\triangle V = 0.02484m^3$

The volume will expand by $0.02484m^3$

7 0

2 years ago

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$