A pitcher can throw a fastball that reaches home plate at 95 mph. What is this speed in m/s

The position vector of a particle of mass 1.65 kg as a function of time is given by = (6.00 î + 4.15 t ĵ), where is in meters an

SashulF [63]

Answer:

L = 41.09 Kg m2 / s The angular momentum does not depend on the time

Explanation:

The definition of angular momentum is

L = r x p

Where blacks indicate vectors

Let's apply this definition our case. Linear momentum

p = m v

Let's replace

L = m r x v

The given function is

x = 6.00 i ^ + 4.15 t j ^

We look for speed

v = dx / dt

v = 0 + 4.15 j ^

To evaluate the angular momentum one of the best ways is to use determinants

$L = m \left[\begin{array}{ccc}i&j&k\\6&4.15t&0\\0&4.15&0\end{array}\right]$

L = m 6 4.15 k ^

The other products give zero

Let's calculate

L = 1.65 6 4.15 k ^

L = 41.09 Kg m2 / s

The angular momentum does not depend on the time

7 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

24- Which of the fallowing devices make use a lens

mezya [45]

Answer:

d.

Explanation:

all because the projector is a lens and the magnifying also the microscope

6 0

2 years ago

Read 2 more answers

HELP ME

Paladinen [302]

Answer: All apply

The periodic table is an arrangement of the chemical elements in the form of a table, ordered by:

-Their atomic number (number of protons)

-Their configuration of electrons

-Their chemical properties

It was progressively developed over time as the scientific knowledge advanced; for this reason many modifications and corrections might be done in the future.

Its usefulness lies in the fact that it allows the existing elements to be organized in a more structured and coherent way, according to the chemical properties they possess. Dividing the table into rows and columns, which represent the periods and groups or families.

Then, with the location and classification of an element according to its group, we can determine how it acts by knowing its chemical and physical characteristics.

This is how with this configuration can be distinguished 4 sets of chemical elements, according to the ease of their atoms to lose or gain electrons, transforming into ions: metals, semimetals, non-metals and noble gases.

This has helped to predict the existence of various elements that have not yet been discovered, because by elements already located in the table and the periodicity found, <u>there are still empty spaces that indicate the composition of the element that has not yet been found</u>.

In addition, this table helps to simplify in some way the teaching of chemical elements and facilitates their learning, as well as their usage in the development of technological innovations.

6 0

2 years ago

What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C

H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ

Heat energy required to change water from 100°C to steam at 100°C

H₄ = mL = 1000 x 2257 = 2257 kJ

Total heat energy required

H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0

3 years ago