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3 years ago

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What is the diameter of a 12lb shot if the specific gravity is of the shot iron in the shot is 6.8, the density of fresh water 6

2.4lb/ft

1 answer:

marysya [2.9K]3 years ago

7 0

Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

$V = \dfrac{4}{3}\pi r^3\ ft^3$

The density of shot is

Using formula of density

$\rho = \dfrac{m}{V}$

Put the value into the formula

$\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}$

We need to calculate the radius

Using formula of specific gravity

$specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}$

Put the value into the formula

$6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}$

$r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}$

$r^3=0.0067514$

$r =(0.0067514)^{\frac{1}{3}}$

$r=0.1890\ ft$

The diameter will be

$d = 2\times r$

$d =2\times0.1890$

$d =0.378\ ft$

Hence, The diameter is 0.378 ft.

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Two 5.0-cm-diameter conducting spheres are 8.0 m apart, and each carries 0.12 mC. Determine (a) the potential on each sphere, (b

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Answer:

Explanation:

Two spheres 10m apart

Each charge on the sphere is 0.12mC= 0.12×10^-3C

Given that the diameter is

5cm=0.05m

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r=d/2

r=0.05/2

r=0.025m

Potential is given as

V=kq/r

k=9×10^9Nm2/C2

a. Potential on each sphere surface.

They are going to have the same value since the sphere are identical

At the surface of the sphere,

r= 0.025m

V=kq/r

V=9E9×0.12E-3/0.025

V=4.32E7Volts

V=4.32×10^7Volts

b. The electric field at the surface of each sphere will be the same since the charge are identical,

So, Electric field is given as

E=kq/r^2

At the surface

E=9E9×0.12E-3/0.025^2

E=1.728×10^8N/C

c. The potential mid way between the two sphere

The potential difference due to the first sphere

The sphere are 10m apart then the distance mid way is 5m

Then, the radius of the sphere is 0.025m

Total distance from the center is 5.025m

Then,

V=kq/r

V1=9E9×0.12E-3/5.025

V1=2.149×10^5 Volts

Potential difference due to the second sphere is the same as the first, since both are identical

Total potential is V1 +V2

V=2.149E5+2.149E5

V=4.3E5Volts

Total potential at the middle due to the two sphere is 4.3×10^5Volts

d. The potential difference between the sphere at any point is equal to the potential difference found at c

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V=4.3×10^5Volts

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if the current in a wire is 2.0 amperes and the potential difference across the wire is 10 volts what is the resistance of the w

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Answer:

R = 2Ω

Explanation:

Potential difference (V) = current (I) * Resistance (R)

V = IR

I = 2.0A

V = 10v

R = ?

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R = V / I

R = 10 / 2

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3 years ago

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Calculating the acceleration of two people pulling a box with same force and angle.

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Here as we can see there we be net horizontal force acting on it

$F_x = F_1cos20 + F_2cos20$

also we know that

$F_1 = F_2 = 80 N$

now we will have

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$F_x = 150.35 N$

now for net force of box we know

$F_x - F_f = ma$

$150.35 - 100 = ma$

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In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one domi

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Answer:

In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one dominates. Make some specific examples of resistorsin parallel and in series to explain this rule of thumb.

Explanation:

Lets start with Resistors in series:

Suppose a 12v battery is connected in a circuit with 3 resistors in series.

Voltage = 12 V

R1 = 1 ohm

R2 = 6 ohm

R3 = 13 ohm

Total resistance is simply sum of all resistors connected in series as following:

Rs = R1 + R2 + R3

Rs = 1 + 6 + 13

Rs = 20 ohm

Here Rs = Total Resistance in series circuit

Now total current can be found by using ohm's law V = IR

I = V/R

Here R = R total

I = 12/20 = 0.6 Ampere

In series circuit current remains same.

so I1 = I2 = I3 = I

Now we can find Power dissipation across every resistor to show the dominant Resistor as:

P1 = I² R1 = (0.6 A)² ( 1 ohm) = 0.36 watt. (a)

P2 = I² R2 = (0.6 A)² ( 6 ohm) = 2.16 watt. (b)

P3 = I² R3 = (0.6 A)² ( 13 ohm) = 4.68 watt. (c)

From equation a, b and c we can conclude that more power is dissipated across R3 which is the Larger than both other resistors R1 and R2.

Now lets take the case of Parallel Circuit with same Values of Voltage and resistors But now in parallel.

In parallel circuit Total resistance is calculated as following:

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/1 + 1/6 + 1/13 = 1 + 0.1667 + 0.0769

1/Rp = 1.2436

Rp = 0.8041 ohm

Now total current is as following:

I = V/R

I = 12/0.8041 = 14.92 A

Now we calculate individual currents :

In parallel circuit Voltage remains same.

I1 = V/R1

I1 = 12/1 = 12 A

Similarly,

I2 = V/R2

I2 = 12/ 6 = 2 A

and

I3 = V/R3

I3 = 12/13 = 0.92 A

Now Power dissipation can be calculated as :

P1 = I1² * R1

P1 = 12² * 1 = 144 * 1

P1 = 144 Watt (1)

Similarly,

P2 = I2² * R2

P2 = 2² * 6 = 4 * 6

P2 = 24 Watt (2)

And

P3 = I3² * R3

P3 = (0.92)² * 13 = 0.8464 * 13

P3 = 11 Watt. (3)

Now from Equation 1, 2 and 3 we can conclude that More power is dissipated across Lower resistor .

Hence in a circuit with parallel resistors, the smaller resistance dominates; in a circuit with resistors in series, the larger one dominates.

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3 years ago