List two examples of energy transformation necessary for the sporting event to take place.

A balloon is rubbed on someone's hair, then placed next to a stream of water. The balloon has an overall negative charge. The wa

sergiy2304 [10]

Answer:

If I'm right it bends away

Explanation:

I learned this in class before it was fun to me

8 0

3 years ago

A 1000 kg car moves with a constant speed 20.0 m/s

skad [1K]

Answer:

13.33 m/s^2

Explanation:

Velocity^2 then divide that by the radius

6 0

3 years ago

A standard baseball has a circumference of apoximately 23cm. If a baseball had the same mass per unit volume as a neutron or a p

BigorU [14]

Baseball, neutrons, and protons are all perfect spheres.

Circumference of the baseball, $L_{\text {ball }}=23 \mathrm{~cm}$ .

mass of the proton or neutron, $m_{\mathrm{p}}=10^{-27} \mathrm{~kg}$

diameter of the proton or neutron, $r_{0}=10^{-15} \mathrm{~m}$

Mass per unit volume is the definition of a material's density. The density equation is

$\rho=\frac{M}{V} .$

Here, M is the mass of the material, V is the volume, and $\rho$ is the density of the material.

We know that formula for the volume of a sphere having diameter d is $V=\frac{\pi d^{3}}{6}$.$

You may obtain this result by plugging this number into the proton/neutron density equation:

$\rho_{0}=\frac{m_{0}}{\left(\frac{\pi d_{0}^{3}}{6}\right)}$

By replacing the variables in the aforementioned equation with their values, you will obtain

$\begin{aligned}\rho_{\circ} &=\frac{10^{-27}}{\left(\frac{\pi\left(10^{-15}\right)^{3}}{6}\right)} \\&=1.91 \times 10^{18} \mathrm{kgm}^{-3}\end{aligned}$

The formula for calculating a circle's circumference with diameter d is $L=\pi d$.$

Using this, you get the diameter of the baseball:

$\begin{aligned}d_{\text {ball }} &=\frac{L_{\text {ball }}}{\pi} \\&=\frac{23}{\pi} \\&=7.32 \mathrm{~cm} \\&=0.0732 \mathrm{~m}\end{aligned}$

The volume of the baseball may be calculated using the formula for the volume of a sphere:

$\begin{aligned}V_{\text {ball }} &=\frac{\pi d_{\text {ball }}^{3}}{6} \\&=\frac{\pi \times(0.0732)^{3}}{6} \\&=2.054 \times 10^{-4} \mathrm{~m}^{3}\end{aligned}$

The mass of the baseball is determined by the concept of density if you suppose that its density is equal to that of a proton or neutron.

$m_{\text {ball }}=\rho_{\circ} \cdot V_{\text {ball }}$

Here, $\rho_{\circ}$ is the density of the proton/neutron.

By replacing the variables in the aforementioned equation with their values, you will obtain

$\begin{aligned}m_{\text {ball }} &=\left(1.91 \times 10^{18}\right) \cdot\left(2.054 \times 10^{-4}\right) \\&=3.92 \times 10^{14} \mathrm{~kg}\end{aligned}$

Therefore, if the density of the baseball was equal to that of the neutron/proton, then its mass would be $3.92 \times 10^{14} \mathrm{~kg}$.$

Learn more about density brainly.com/question/15164682

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7 0

1 year ago

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

lukranit [14]

Answer:

$\frac{dA}{dt} = 28800 \ m^2/year$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Area of a Rectangle: A = lw

Algebra I

Exponential Property: $w^n \cdot w^m = w^{n + m}$

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

$\frac{dw}{dt} = 24 \ m/year$

Step 2: Rewrite Equation

Substitute in l: A = (2w)w
Multiply: A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

Differentiate [Basic Power Rule]: $\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}$
Simplify: $\frac{dA}{dt} = 4w\frac{dw}{dt}$

Step 4: Find Rate

Use defined variables

Substitute: $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = 28800 \ m^2/year$

3 0

3 years ago

Read 2 more answers

A pan of ice is placed over a flame.

AveGali [126]

A. The molecules start packed together very tightly in a solid. Then when it turns to water, the molecules can move around each other freely, but still contained. When water turns to vapor, the molecules are going crazy moving around. They are not contained at all and bounce of of each other freely.

b. The temperature rises. (ice turns to water at 33 degrees and water turns to vapor at 212 degrees)

8 0

3 years ago