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2 years ago

Help!!! Please need it fast. If correct I’ll give b

Physics

2 answers:

balu736 [363]2 years ago

5 0

Alleles form when 2 parent produce 1.

Lynna [10]2 years ago

3 0

Answer:

if bother parents have brown eyes but they carry the allele for blue eyes a quarter of the kids will have blue eyes and three quarters will have brown eyes. hope this help I'm sorry if it doesn't.

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What is the density of a piece of wood with a mass of 25 kg<br> and a volume of 0.0385 m³?

Orlov [11]

Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

$p = \frac{mass \: (in \: kg)}{volume \: (in \: {m}^{3}) }$

We can substitute the givenmass and volume to find density of the object.

$p = \frac{25kg}{0.0385 {m}^{3} } \\ = 649kg \: per \: {m}^{3}$

Therefore the density of this object is 649kg/m^3.

7 0

1 year ago

Magnesium is a shiny, flexible metal that can be burned in the presence of air.

alukav5142 [94]

Answer:

A new substance was formed

Explanation:

According to this question, a shiny and flexible metal called Magnesium (Mg) is burnt in air to produce a white powder that has no shiny or flexible properties, however, has more weight than the magnesium metal itself.

This is possible because a CHEMICAL CHANGE has occured, hence, a new substance has been formed. The formation of a new substance during the burning process (chemical reaction), induced the increase in mass.

8 0

2 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

4 0

3 years ago

How is energy transferred throughout the roller coaster

solmaris [256]

Answer:

On a roller coaster, energy changes from potential to kinetic and back again many times over and over the course of the ride. Kinetic energy is energy that an object has because of its motion. All moving objects possess kinetic energy, which is determined by the mass and speed of the object.

Explanation:

3 0

3 years ago

Read 2 more answers

Why can’t a real machine ever have 100% efficiency

Harman [31]

Answer:

Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.

4 0

3 years ago