A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first

Answer:

The force will rise in direct relation to the mass of the objects

Explanation:

The gravitational acceleration remains constant. It is measured in meters per second squared or m/s². The average value is 9.81 m/s², calculated from observations made on varying surfaces. In reality, the acceleration can vary based on the geographical shape of the Earth relative to the earth's magnetic field and gravitational force.

For instance, if a single washer weighs 20 kg, with the gravity at 9.81 m/s², the weight would be:

F = ma

  =

If there are three washers, the total weight calculates as:

F = 3 * 20 * 9.81

  = 588.6 N

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

Therefore, its weight in air is noted as

Next, the buoyant force acting on the lead is defined as

We know that

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

Answer:

Explanation:

Utilizing the formula for surface radiation P (energy per second in Watts) = emissivity constant * surface area * Stefan-Boltzmann constant * Temperature in Kelvin^4 *

2.7*10^31 = 1* 5.67*10^-8*A*11000^4

Rearranging to solve for A = 2.7*10^31 / (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2

Assuming the shape is spherical, the surface area is = 4πR^2 (radius of Rigel)

R = √(0.3261*10^23 / 4*π) = 5.1 * 10^10m

B) repeating the same calculation

2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of Procyon

Rearranging gives A = 2.1*10^23/(5.67*10^-8*10^16)

A = 0.37*10^15

Assuming the star is spherical;

A = 4πR^2 where R is Procyon's radius

R = √(0.37*10^15/4π) = 5.4*10^6m