Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work

Superman is standing 393 m horizontally away from Lois Lane. A villain drops a rock from 4.00 m directly above Lois. If Superman

Sergio039 [100]

Answer:

-963.93 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4\times 2}{9.81}}\\\Rightarrow t=0.903\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 393=0\times 0.0903+\frac{1}{2}\times a\times 0.903^2\\\Rightarrow a=\frac{393\times 2}{0.903^2}\\\Rightarrow a=963.93\ m/s^2$

The acceleration of Superman would be -963.93 m/s² from Lois' perspective

6 0

2 years ago

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

#2

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

2 years ago

The connections of many simple pieces in the brain is evidence of the:

olga55 [171]

Brian’s Complexity Brian’s Complexity Brian’s Complexity Brian’s Complexity

6 0

2 years ago

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Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a so

Citrus2011 [14]

Answer:

The force does the ceiling exert on the hook is 269.59 N

Explanation:

Applying the second Newton law:

F = m*a

From the attached diagram, the net force in object 1 is:

$m_{1} a=T_{1} -W_{1}$

In object 2:

$m_{2} a=W_{2} -T_{2}$

Adding the two equations:

$m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2} -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a$ (eq. 1)

The torque:

$\tau =I\alpha$

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

$a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}$

Torque due the tension is equal:

$\tau =r(T_{2} -T_{1} )$

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

$a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }$

Where

W₁ = 75 N

W₂ = 125 N

W = 80 N

$a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}$

The net force is:

$F_{n} =F-W-T_{1} -T_{2}\\0=F-W-W_{1} (\frac{a}{g} +1)-W_{2} (1-\frac{a}{g})\\F=W+W_{1} +W_{2} +\frac{a}{g} (W_{1} -W_{2} )\\F=80+75+125+\frac{2.04}{9.8} (75-125)\\F=269.59N$

4 0

2 years ago

To learn to apply the concept of current density and microscopic Ohm's law. A "gauge 8" jumper cable has a diameter d of 0.326 c

alukav5142 [94]

Answer:

Resistivity = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Option A is correct. Copper.

Explanation:

Given,

I = 30.0 A

d = 0.326 cm = 0.00326 m

E = 0.062 V/m

The current density for the wire is given as

J = (I/A)

where J = current density = ?

I = current = 30.0 A

A = Cross sectional Area of the wire = (πd²/4) = [π×(0.00326²)/4] = 0.0000083503 m²

J = (30 ÷ 0.0000083503) = 3,592,685.3 A/m²

On a microscopic scale, Ohm's law can be stated as

E = Jρ

where E = Electric field = 0.062 V

J = Current density = 3,592,685.3 A/m²

ρ = Resistivity of the material = ?

ρ = (E/J)

ρ = 0.062 ÷ 3,592,685.3 = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Hope this Helps!!!

8 0

2 years ago

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