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2 years ago

Which possesses the most gravitational potential energy?

Physics

2 answers:

ser-zykov [4K]2 years ago

8 0

Answer:

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shtirl [24]2 years ago

4 0

I’m think it’s B or A

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A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point

pogonyaev

Answer:

L = 1.545 m

Explanation:

Let the total length of the rod is L

now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum

so we will have

$mg \times d = F(L - d)$

so we have

$325\times 9.8 \times 0.266 = F(L - 0.266)$

F = 663 N

$848 = 663(L - 0.266)$

$L = 1.545 m$

6 0

3 years ago

Differences between concave and convex meniscus

BARSIC [14]

Explanation:

A concave meniscus,(normally seen) occurs when the molecules of the liquid are attracted to those of the container. This occurs with water and a glass tube. A convex meniscus occurs when the molecules have a stronger attraction to each other than to the container, as with mercury and glass.

4 0

3 years ago

Read 2 more answers

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goblinko [34]

Roygbv evjnefvvnefv ekfv k kn ke nv

5 0

2 years ago

A biconvex lens is formed by using a piece of plastic(n=1.70).

creativ13 [48]

Answer:

$f =17.15\ cm$

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

Radius of curvature of the back surface, R₂ = 30 cm

focal length= ?

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

R₁ = +20 cm

R₂ = -30 cm

n = 1.70

$\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})$

$\dfrac{1}{f}=0.70 \times 0.0833$

$f = \dfrac{1}{0.7 \times 0.0833}$

$f =17.15\ cm$

the focal length of the lens is equal to 17.15 cm

3 0

3 years ago

Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What

yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0

2 years ago