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AfilCa [17]
4 years ago
11

Mass = 567.5 g Volume = 50.0 cm3 What is the density of lead? 1.135 g/cm3 11.35 g/cm3 113.5 g/cm3 1,135 g/cm3

Physics
1 answer:
nlexa [21]4 years ago
8 0
Density = mass / volume So density =567.5/50 = 11.35 g/cm3.
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4 years ago
Sally Sizzle builds a circuit by connecting three resistors in oarallel with a 9v battery. The current in this curvuit is 3.6 am
Sergeeva-Olga [200]

Answer:

The new resistor has resistance of 9Ω.

Explanation:

The effective resistance of three resistance is

$ (1). \: \: \frac{1}{R_e}  = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}   $.

we know the the circuit with 9V battery has current of 3.6A; therefore, R_e must be

V = IR_e

R_e = \dfrac{V}{I} =  \dfrac{9V}{3.6A}

R_e = 2.5\Omega,

which means

$\frac{1}{2.5\Omega }  = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}   $

$\boxed{0.4\Omega^{-1} = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}  .} $

Now, Sally wants to connect another resistor in parallel so that the current becomes 4.6A. The resistance required for that is

R_{new} = \dfrac{9V}{4.6A}

R_{new} = 1.957 \Omega.

which means the 4th resistor R_4 must satisfy the equation

$ (2), \: \: \frac{1}{1.957\Omega}  = \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3}+\frac{1}{R_4} ,   $

and since

$ \frac{1}{R_1}+ \frac{1}{R_2}+ \frac{1}{R_3} =0.4\Omega^{-1},  $

equation (2) becomes

$ \frac{1}{1.957\Omega}  =0.4\Omega^{-1}+\frac{1}{R_4} ,   $

which we solve for R_4:

$ 0.511\Omega^{-1} =0.4\Omega^{-1}+\frac{1}{R_4} ,   $

0.111\Omega^{-1} = \dfrac{1}{R_4}

\boxed{R_4 = 9\Omega}

Thus, the resistance of the new resistor is 9 ohms.

3 0
4 years ago
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swat32

Answer:

the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour.

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