Which element is malleable and ductile

Metals in group 2 on the period table most commonly form which type

LuckyWell [14K]

Answer:

alkaline earth metals

Group 2 metals, the alkaline earth metals, have 2 valence electrons, and thus form M2+ ions. The halogens, Group 17 , reach a full valence shell upon reduction, and thus form X− ions

Explanation:

6 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

4 years ago

Identify some of the effects of urbanization

dangina [55]

Answer: Heyyyyyyyyyyyyyy

here you goo

"some of the effects of urbanization"

Poor air and water quality, insufficient water availability, waste-disposal problems, and high energy consumption are exacerbated by the increasing population density and demands of urban environments

"hope this helps"

4 0

4 years ago

It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atm

klio [65]

Answer:

a) mb = 0.0596 kg ; r = 0.974 m

b) a = 754 m/s^2 .. (Upward)

c) mL = 5.96 kg

Explanation:

Given:-

- The density of Mars atmosphere , ρ = 0.0154 kg/m^3

- The surface density of ballon, σ = 5.0g/m^2

Solution:-

(a) What should be the radius and mass of these balloons so they just hover above the surface of Mars?

- We will first isolate a balloon in the Mar's atmosphere and consider the forces acting on the balloon. We have two forces acting on the balloon.

- The weight of the balloon - "W" - i.e ( Tough plastic weight + Gas inside balloon). Since, the balloon is filled with a very light gas we will assume the weight due to gas inside to be negligible. So we have:

W = mb*g

Where, mb: Mass of balloon

g: Gravitational constant for Mars

- The mass of the balloon can be determined by using the surface density of the tough plastic given as "σ" and assuming the balloon takes a spherical shape when inflated with surface area "As".

As = 4πr^2

Where, r: The radius of balloon

So, mb = 4σπr^2

- Substitute the mass of balloon "mb" in the expression developed for weight of the balloon:

W = 4*σ*g*πr^2 ......... Eq1

- The weight of the balloon is combated by the buoyant force - "Fb" produced by the volume of Mars atmosphere displaced by the balloon acting in the upward direction:

Fb = ρ*Vs*g

Where, Vs : Volume of sphere = 4/3 πr^3

So, Fb = ρ*g*4/3 πr^3 ....... Eq 2

- Apply the Newton's equilibrium conditions on the balloon in the vertical direction:

Fb - W = 0

Fb = W

ρ*g*4/3 πr^3 = 4*σ*g*πr^2

r = 3σ / ρ

r = 3*0.005 / 0.0154

r = 0.974 m .... Answer

- Use the value of radius "r" and compute the "mb":

mb = 4σπr^2

mb = 4*0.005*π (0.974)^2

mb = 0.0596 kg ... Answer

(b) If we released one of the balloons from part (a) on earth, where the atmospheric density ρ = 1.20kg/m^3, what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down?

- The similar analysis is to be applied when the balloon of the same size i.e r = 0.974 m and mass mb = 0.0596 kg is inflated on earth with density ρ = 1.20kg/m^3.

- Now see that the buoyant force acting on the balloon due to earth's atmosphere is different from that found on Mars. So the new buoyant force Fb using Eq2 is:

Fb = ρ*g*4/3 πr^3

Where, g: Gravitational constant on earth = 9.81 m/s^2

Fb = (1.20)*(9.81)*(4/3)* π*(0.974)^3

Fb = 45.5 N

- Apply the Newton's second law of motion in the vertical direction on the balloon:

Fb - W = mb*a

Where, a: The acceleration of balloon

a = (Fb - W) / mb

a = Fb/mb - g

a = 45.5/0.0596 - 9.81

a = 754 m/s^2 (upward) ..... Answer

c), d) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

- The new radius of the balloon - "R" -is five times what was calculated in part (a):

- Apply the Newton's equilibrium conditions in the vertical direction on the balloon with the addition of downward weight of load "WL":

Fb - W - WL = 0

WL = Fb - W

mL*g = ρ*g*4/3 πR^3 - 4*σ*g*πR^2

Where, mL : The mass of load due to instrument package

mL = ρ*4/3 πR^3 - 4*σ*πR^2

mL = 0.0154*4/3*π*(5*0.974)^3 - 4*(0.005)*π*(5*0.974)^2

mL = 7.45 - 1.45

mL = 5.96 kg ..... Answer

6 0

3 years ago

Read 2 more answers

As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.

Ganezh [65]

Answer: (a) The bicycle is ahead of the car for 4 s.

(b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

$v=u+at$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

$22.3=0+4.02t$

$t=\frac{22.3}{4.02}$

t= 5.5 s

Use the equation of the motion to calculate the time taken by the bicycle.

$v=u+at_{1}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

$8.94=0+5.81t_{1}$

$t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s$

Calculate the time interval for which the bicycle is ahead of the car.

$t-t_{1}= 5.5 s - 1.5s$

$t-t_{1}= 4s$

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

$S=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

$S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}$

$S= 60.8 m$

Use the equation motion to calculate the distance covered by the bicycle.

$S_{1}=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

$S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}$

$S_{1}= 5.8 m$

Calculate the maximum distance covered by the bicycle to lead the car.

$S-S_{1}=60.8-5.8=55m$

Therefore, the bicycle leads the car by the maximum distance of 55 m.

6 0

3 years ago