A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of

Individuals inherit one factor per trait from each parent, and a dominant factor can conceal the expression of a recessive one when both are present.

<span>Answer
A person who weighs 220 lb has less mass than someone who weighs 288 lb, so accelerating the 220 lb player requires less force. The heavier player therefore carries greater momentum. Because 288 lb corresponds to more weight (and mass), that player has higher inertia and is harder to stop. For these reasons it is easier to tackle a 220 lb player than a 288 lb player. 
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Q should be positioned 4.8 miles east of point B. Explanation: From the diagram, we can define the construction cost as a function of angle θ (as illustrated). The underwater pipe length (marked in blue) equals 6/cos θ, while the land pipe length (marked in brown) is (8 - 6*tan θ). The total construction cost formula is: Construction Cost = (6/cos θ)(6000) + (8 - 6*tan θ)(3750). This formula is represented in terms of θ, which can vary from 0 degrees to 53.13 degrees according to the diagram. To find the angle θ that minimizes the construction cost, we differentiate the Construction Cost function with respect to θ and set it to zero. The derivative yields: -4500(5*sec θ – 8*tan θ)(sec θ) = 0, leading to θ = 38.68 degrees. By substituting θ, we can determine the distance of Q from B, which equals 6*tan θ. This calculates to a distance of 4.8 miles.

Answer:

A) 5.1*10^10m B) 5.4*10^6m

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

Answer:

The final size is nearly the same as the initial size because the increase in size is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass,

Initial size of the wave packet,

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

This proton energy is

Thus, the speed of the proton, v

The time to cover 1 km = 1000 m of distance is calculated as:

T =

T =

According to the dispersion factor;

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

where

= final width