a professional baseball player can pitch a baseball with a velocity of 44.7m/s towards home plate. If a baseball weighs 1.4 N, h

Response:

A protractor to gauge the angle between the inclined plane and the horizontal

Explanation:

The student must elevate the free end of the adjustable inclined plane until the object just begins to slide and record the angle at that precise moment. At this juncture, the frictional force is balanced by the weight component aligned with the incline. That is:

and  

Consequently, the coefficient of static friction can be entirely established by calculating the tangent of the angle formed by the incline with the horizontal.

For this, the sole additional tool needed is a protractor for angle measurement.

According to Snell’s Law: Where: is the index of refraction of the first medium (glass), and is the index of refraction of the second medium (ice). The angle of incidence and the angle of refraction are represented by and. The refractive index quantifies the speed of light in a medium. The critical angle is identified as the angle at which total internal reflection occurs, meaning no light passes through into another medium. This phenomenon happens only when the light is transitioning from a medium with a higher index of refraction to one with a lower index of refraction.

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

Answer:

0.6

Explanation:

The formula for the volume of a sphere is

Thus

The radius of the disk is

Applying angular momentum conservation;

The of the sphere =

of the disk =

= 0.6

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We utilize the formula

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the angular momentum magnitude of the person 2 meters prior to jumping onto the merry-go-round?

The equation we apply is

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the angular momentum of the person just before she hops onto the merry-go-round?

We utilize the formula

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular velocity of the merry-go-round after the individual jumps on?

We can apply the Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒  I₀*ω₀ = (I₀ + m*R²)*ω₁

At this point, we can determine ω₁

⇒  ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒  ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒  ω₁ = 0.769 rad/s

5) Once the merry-go-round moves at this new angular speed, what force must the person exert to hold on?

We must calculate the centripetal force as follows

Fc = m*ω²*R  

⇒  Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) When the person is halfway around, they choose to simply release their hold on the merry-go-round to exit the ride.

What is the linear speed of the person the moment they exit the merry-go-round?

we can apply the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular velocity of the merry-go-round after the individual releases their hold?

ω₀ = 1.53 rad/s

It returns to its original angular speed