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3 years ago

6.P.2.1 Matter: Properties and Changes Notes

Physics

1 answer:

Zinaida [17]3 years ago

8 0

Answer:

The __law__ conservation

of state that either energy may produced nor destroyed

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force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it

Yanka [14]

Answer:

Given force=10lb

L1=4in converting to feet

But 0.08333ft= 1 inch

Then 4 inch is 0.3332

6inch is 0.49998

But hookes law states

F=Kx where F is force,K is the force constant ,X

K=F/X=10/0.3333=30N/m

Integrating this

Integral of 30x with limit 0.333 to 0.5

F=30x^2/2=15x^2substing the limit

F=(15(0.5^2-0.33^2)=2.08ft-lb

Explanation:

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3 years ago

Direct current made it possible to distribute electric power over greater areas.

Mademuasel [1]

True, I'm not the best when it comes to science, but I'm pretty sure it's this

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3 years ago

A semi-infinite solid is

Anarel [89]

Answer:

D. infinitely extended in all directions

Explanation:

A semi infinite solid is infinitely extended in every direction. It has a single surface and can extend when heat is applied.

The body of a semi infinite solid is idealised, that is, when there is heat present, it expands in all directions to infinity. It can be used for a thick wall because its shape can be changed when subjected to different levels of heat near its surface.

It is also expands as heat is applied because its thickness is negligible.

This idealized body is used for earth, thick wall, steel piece of any shaped quenched rapidly etc indetermining variation of temperature near its surface & other surface being too far to have any impact on the region in short period of time since heat doesn’t have sufficient time to penetrate deep into body thus thickness can be neglected

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3 years ago

Consider the circuit below, which is powered by a 6-v battery. switch s is opened at t = 0 after having been closed for a long t

Semmy [17]

The answer should be B

3 0

3 years ago

Urgent!

Masja [62]

mass of iron block given as

$m_1 = 1.90 kg$

density of iron block is

$\rho = 7860 kg/m^3$

now the volume of the iron piece is given as

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

$V = \frac{1.90}{7860} = 2.42* 10^{-4} m^3$

Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as

$F_b = \rho_L V g$

here we know that

$\rho_L$ = density of liquid = 916 kg/m^3

$F_b = 916* 2.42 * 10^{-4} * 9.8$

$F_b = 2.17 N$

Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium

$F_s + F_b = mg$

$F_s + 2.17 = 1.90* 9.8$

$F_s = 16.45 N$

So reading of spring balance will be 16.45 N

Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block

$F_n = F_g + F_b$

$F_n = (1 + 2.50)*9.8 + 2.17$

$F_n = 34.3 + 2.17 = 36.47 N$

So the other scale will read 36.47 N

3 0

3 years ago