Which statements describe why organisms are classified? Check all that apply.

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she

tester [92]

Answer:

a) I = 1,75 10-² kg m² and b) I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

I = I core + I shell

The moment of inertia of a solid sphere is

I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

R = d / 2

R= 0.196 m / 2 = 0.098 m

I core = 2/5 1.6 0.098²

I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

R = 0.206 / 2

R = 0.103 m

I shell = 2/3 1.6 0.103²

I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

I = I core + I shell

I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

I = 17.47 10⁻³ kg m²

I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

R = 0.216 / 2

R = 0.108 m

It is a uniform sphere

I = 2/5 M R²

I = 2/5 3.2 0.108²

I = 1.49 10⁻² kg m²

7 0

3 years ago

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

Why are we integrating the differential x-component of the e field of this charged ring?

iragen [17]

Please give us the picture and tell us what quantity you're asked to find.

6 0

3 years ago

2. A car traveling 300 miles in 5 hours is an example of

Tom [10]

The answer would be a speed

5 0

3 years ago

Describe the difference between renewable and nonrenewable resources. Give at least two advantages and two disadvantages of each

elixir [45]

Answer: (A), (C), and (E)

Explanation:

5 0

3 years ago