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4 years ago

6

As the particles of a solid undergo sublimation, they

1 answer:

dolphi86 [110]4 years ago

5 0

D) Move farther apart from one another. Sublimation is when a solid changes to a gaseous state. The particles that were tightly compact into a solid are released and spread away from eachother into the air as a gas.

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The radius of curvature of a spherical mirror is 20cm.What is its focal length?

hichkok12 [17]

Answer:

$\boxed{\sf Focal \ length = 10 \ cm}$

Given:

Radius of curvature (R) of a spherical mirror = 20

To Find:

Focal length (f)

Explanation:

Formula:

$\boxed{ \bold{\sf Focal \ length \ (f) = \frac{Radius \ of \ curvature \ (R)}{2}}}$

Substituting value of R in the equation:

$\sf \implies f = \frac{20}{2}$

$\sf \implies f = \frac{ \cancel{2} \times 10}{ \cancel{2}}$

$\sf \implies f = 10 \: cm$

5 0

4 years ago

Read 2 more answers

If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass

EleoNora [17]

Simply, apply the formula $E = \frac{1}{2}mv^{2}$ and insert the values of m = mass, v = velocity and E = Energy.
The result will be $72J = \frac{1}{2}m(12m/s)^{2}$ , m = 1 kg

5 0

3 years ago

A rectangular piece of cardboard, whose area is 352 square centimeters, is made into an open box by cutting a 2-centimeter sq

Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

$xy=352$

$y=\dfrac{352}{x}$ -------- (1)

The volume of the rectangular piece

$2(x-4)(\dfrac{352}{x}-4)=432$

$x^2-38x+352=0$

$(x-16)(x-22)=0$

x=16,22

Put the value of x in the equation (I)

For x = 16

$y=\dfrac{352}{16}=22$

For x = 22

$y=\dfrac{352}{22}=16$

Dimension of cardboard is 22 m by 16 m

8 0

4 years ago

A sort of "projectile launcher" is shown below. A large current moves in a closed loop composed of fixed rails, a power supply,

Zarrin [17]

Answer:

wallah i don't understand anything with my stoopid brain

Explanation:

3 0

3 years ago

During a compression at a constant pressure of 290 Pa, the volume of an ideal gas decreases from 0.62 m3 to 0.21 m3. The initial

Aloiza [94]

Answer:

a) -41.1 Joule

b) 108.38 Kelvin

Explanation:

Pressure = P = 290 Pa

Initial volume of gas = V₁ = 0.62 m³

Final volume of gas = V₂ = 0.21 m³

Initial temperature of gas = T₁ = 320 K

Heat loss = Q = -160 J

Work done = PΔV

⇒Work done = 290×(0.21-0.62)

⇒Work done = -118.9 J

a) Change in internal energy = Heat - Work

ΔU = -160 -(-118.9)

⇒ΔU = -41.1 J

∴ Change in internal energy is -41.1 J

b) V₁/V₂ = T₁/T₂

⇒T₂ = T₁V₂/V₁

⇒T₂ = 320×0.21/0.62

⇒T₂ = 108.38 K

∴ Final temperature of the gas is 108.38 Kelvin

5 0

4 years ago