If aluminum loses all 3 electrons how do you write the ion

In the park, there are 25 pigeons, 15 squirrels, 5 rabbits, and 5 stray cats. Why

Debora [2.8K]

Answer: Line graph should be used to show how one variable changes over time not to show multiple categories or variables are at one specific point in time.

Explanation:

In maths, statistics, and related fields, graphs are used to visually display variables and their values. In the case of line graphs, these are mainly used to display evolution or change of a variable over time. For example, a line graph can show how the number of divorces changed from 1920 to 2010.

In this context, the number of different animals in the park cannot be represented through a line graph because this situation does not imply a variable changing over time. Moreover, this situation includes multiple variables or categories of animals and the data shows only one specific point in time, which can be better represented through a bar graph.

8 0

3 years ago

What is the new current?

nikdorinn [45]

Voltage = current x resistance
since R is doubled, current must reduce by half.
So,
new current = 120/2 = 60mA

3 0

3 years ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$

Upon substituting our given values, we will get:

$\frac{(T_1)^2}{(T_2)^2}=\frac{(238,000)^3}{(1,222,000)^3}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272000000000}{1824793048000000000}$

$\frac{(T_1)^2}{(T_2)^2}=\frac{13481272}{1824793048}$

$\frac{(T_1)^2}{(T_2)^2}=0.0073878361246365$

Taking square root of both sides, we will get:

$\frac{T_1}{T_2}=\sqrt{0.0073878361246365}$

$\frac{T_1}{T_2}=0.0859525225030452495$

$\frac{T_2}{T_1}=\frac{1}{0.0859525225030452495}$

$\frac{T_2}{T_1}=11.634329870476699\approx 11.634$

$T_2=11.634\cdot T_1$

This implies that time period of Titan about Sturn is 11.634 times more compared to time period of Enceladus about Saturn.

So, basically Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

8 0

3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

6 0

2 years ago

If a tank contains water 4 m deep, what is the pressure at the bottom of the tank? (Neglect the atmospheric pressure.)

makkiz [27]

Given:

Water 4 m deep

Required:

Pressure at the bottom of the tank

Solution:

p2 – p1 = gh

p2 – p1 = p = gh

p = gh = 1000kg/m3 (9.8m/s2)(4m)

6 0

3 years ago