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3 years ago

Mildred takes a pound of frozen hamburger meat out of the freezer and puts it into the refrigerator.

1 answer:

7 0

The meat in the freezer is frozen.
Everything else in the freezer is frozen too.
Nothing in the refrigerator is frozen.
The freezer is colder than the refrigerator. <span>

Mildred takes a pound of frozen hamburger meat out of the freezer
and puts it into the refrigerator. The meat is colder than anything
else that's in there.

Heat flows from the air in the refrigerator into the frozen hamburger (C)
and warms up the meat. When the temperature of the meat warms up
to the temperature of the air in the refrigerator, the heat stops flowing.</span>

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A 20 cm-radius ball is uniformly charged to 71 nC.

artcher [175]

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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3 years ago

How strongly the planet you're on pulls on you is your

ladessa [460]

Is the answer you are looking for Gravity? Gravity is what pulls us down to earth.

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3 years ago

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Define mass .Also write the table showing the relation of 'kg' with it's sub- multiples and multiples.

lys-0071 [83]

Answer:

the quantity of matter contained in a body is called mass .si unit of mass is kilogram.weight is effected by gravity but mass is not effected by gravity.

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2 years ago

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Suppose you lift a 20 kg box by a height of 1.0 m.

9966 [12]

W=mgh W=(20)(9.8)(1) w=196J

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3 years ago

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A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.

ycow [4]

Answer:

σ = 3.402 KPa , γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

Find the shearing stress, shearing strain and shear modulus.

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

- The shear effect results in a stress in the gelatin block.

- The formulation of stress ( σ ) is given below:

σ = F / A

Where,

A : The surface area of the object that experiences the shear force.

- The top surface have the following dimensions:

A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

σ = 49 / 0.0144

σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

γ = x / L

γ = 10 / 40

γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- It is mathematically expressed as a ratio of shear stress ( σ ) and shear strain ( γ ):

G = σ / γ

G = 3.402 / 0.25

G = 13.608 KPa

7 0

3 years ago