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3 years ago

5

How is thermal energy transferred during conduction? Check all that apply.

1 answer:

rodikova [14]3 years ago

3 0

Answer:

2, 3 and 5.

Explanation:

When a solid is heated, the particles vibrate. These vibrations are passed through the solid and energy is transferred easily this way. Metals are good thermal conductors. The particles in solids are close to eachother and when particles are heated the speed up and move faster, which is the reason they vibrate.

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Which statement describe. good hypothesis ? Check all that apply

larisa [96]

A good hypothesis describes your ideas and how you think the experiment will conclude. Additionally, hypotheses have scientific information included.

7 0

3 years ago

If revenue is $3,000 and operating expenses are $4,000, cash flow equals _____. -$1,000 $1,000 -$7,000 $7,000

valkas [14]

Answer:

-$1,000

Explanation:

$3000 - $4000

4 0

3 years ago

19 dm expressed in millimeters

alina1380 [7]

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1 year ago

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

$\dfrac{I_{CM}}{I}=\dfrac{2}{7}$

$I_{CM}=\dfrac{2}{7}I$

the correct answer is option E

4 0

3 years ago

A 1.0-m long wire is carrying a certain amount of current. The wire is placed perpendicular to a magnetic field of strength 0.20

Marysya12 [62]

For a current-carrying wire running perpendicular to a magnetic field, the magnetic force acting on the wire is given by:

F = ILB

F = magnetic force, I = current, L = wire length, B = magnetic field strength

Given values:

F = 0.60N, L = 1.0m, B = 0.20T

Plug in and solve for I:

0.60 = I(1.0)(0.20)

I = 3.0A

5 0

3 years ago