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3 years ago

If the sun is scaled down to 9.5 inches in diameter, the earth would be about _____ inches in diameter

Physics

1 answer:

vovangra [49]3 years ago

7 0

Answer: 0.087 inches

Explanation:

This problem can be solved applying the Rule of three, which is a mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known.

In this sense, the actual diameter of the Sun is 1,391,020 km and the actual diameter of the Earth is 12,742 km.

Now, applying the Rule of three:

Sun 1,391,020 km ------ 9.5 inches

Earth 12,742 km --------- ? inches

$? inches=\frac{(12,742 km)(9.5 inches)}{1,391,020 km}$

$? inches=0.087 inches$

Therefore:

If the sun is scaled down to 9.5 inches in diameter, the earth would be about 0.087 inches in diameter.

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An 800-g block of ice at 0.00°C is resting in a large bath of water at 0.00°C insulated from the environment. After an entropy c

Allisa [31]

Answer:

Unmeltedd ice = 308.109 g

Explanation:

Gibbs Free energy:

A systems Gibbs Free Energy is defined as the free energy of the product of the absolute temperature and the entropy change less than the enthalpy change.

Therefore, G = ΔH-TΔS

where G is Gibbs Free Energy

ΔH is enthalpy change

T is absolute temperature

ΔS is entropy change

Here since there is a phase change, therefore G will be 0.

∴ΔH = TΔS

Given: Temperature, T = 0°C = 273 K

Entropy change,ΔS = 600 J/K

Latent heat of fusion of water = 333 J/g

∴ΔH = TΔS

∴ΔH = 273 x 600

= 163800 J

So this is the amount of enthalpy that will be used into melting of ice.

∴ΔH = mass of ice melted x latent heat of fusion of water

Mass of ice melted = ΔH / latent heat of fusion of water

= 163800 / 333

= 491.891 g

This is the mass of ice melted.

And initial amount of ice is 800 g

Amount of ice left after melting = Initial amount of ice - amount of ice melted

= 800-491.891

= 308.109 g

Amount of ice remained after melting = 308.109 g

8 0

3 years ago

A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

pickupchik [31]

Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

3 0

4 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

3 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

3 years ago

Would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that was

Ber [7]

More compressed. moving up = apparent weight (i.e., your norma force) is greater. this means you’ll weighr more and push those springs down even more than you would at rest.

6 0

3 years ago