A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m

1) The work performed on the spores is

2) The spores land 0.32 m away

Explanation:

1)

According to the work-energy principle, the work done on the spores equals their change in kinetic energy:

where

W is work done

m represents the spore's mass

v is the final velocity

u is the initial velocity

Given:

spore mass =

initial velocity u = 0 (starting from rest)

final velocity v = 7.0 m/s

Calculating the work:

2)

The spores follow projectile motion, moving along a parabolic trajectory made of two motions:

- Constant speed horizontally

- Vertically accelerated due to gravity

Considering vertical motion first, using kinematics:

where

s = 0.01 m (shooting height)

u = 0 (no initial vertical velocity, horizontal ejection)

t = time of flight

g = acceleration due to gravity

Solving for t:

Now for horizontal displacement, when velocity is constant:

where

horizontal velocity =

time t = 0.045 s

Calculating distance d:

Thus, the spores land 0.32 meters away.

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Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by .

                   Here, m denotes the mass of the object

                               v is the velocity or speed of the object

                               r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F =

                               where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F=

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.

Answer:

b = 0.6487 kg / s

Explanation:

In the context of oscillatory motion, friction is related to velocity,

               fr = - b v

where b represents the friction coefficient.

Upon solving the equation, the angular velocity is represented as

               w² = k / m - (b / 2m)²

In this case, we're given an angular frequency w = 1Hz, the mass m = 0.1 kg, and the spring constant k = 5 N / m. This allows us to derive the friction coefficient.

             

Let’s denote

               w₀² = k / m

               w² = w₀² - b² / 4m²

               b² = (w₀² -w²) 4 m²

Now, let's calculate the angular frequencies.

             w₀² = 5 / 0.1

             w₀² = 50

             w = 2π f

             w = 2π 1

             w = 6.2832 rad / s

Substituting values yields

               b² = (50 - 6.2832²) 4 0.1²

               b = √ 0.42086

                b = 0.6487 kg / s

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