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2 years ago

PLZ HELP!! List properties of ions.

1 answer:

3 0

Ionic compounds typically have high melting and boiling points, and are hard and brittle. when melted or dissolved they become highly conductive, because the ions are mobilized.

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How many joules of kinetic energy does a pendulum have when it has 100 joules of potential energy

zloy xaker [14]

Answer:

The maximum kinetic energy is 100 j.

Explanation:

<h3>The kinetic energy = (potential energy) + (kinetic energy) and the potential energy of 0 J implying its kinetic energy is 100 J, which is its maximum. </h3>

4 0

2 years ago

Read 2 more answers

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

2 years ago

Carol drops a stone in a mine shaft 122.5 metres deep.How

Verizon [17]

Answer:

T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken = ?

time taken by the stone to reach at the bottom

using equation of motion

$s = u t + \dfrac{1}{2}gt^2$

initial speed , u = 0 m/s

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 122.5}{9.8}}$

t = 5 s

time taken by the sound to travel

d =v x t

$t = \dfrac{d}{v}$

$t = \dfrac{122.5}{340}$

t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

T = 5.36 s

7 0

2 years ago

a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column

scoray [572]

Answer:

The answer is " $1155\ \frac{kg}{m^3}$ "

Explanation:

Please find the complete question in the attached file.

$p = p_0 + ?gh$

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

$h_1 = 11 \ cm\\\\h_2= 3 \ cm$

The pressures would be proportional to the quantity $11-3 = 8$ cm from below the surface at the interface between both the oil and the liquid.

$\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\$

$= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}$

8 0

2 years ago

Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water a

PolarNik [594]

Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

Explanation:

(a)

$L_{f}$ = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

$Q_{f}$ = Energy required to melt the ice

Energy required to melt the ice is given as

$Q_{f}$ = m $L_{f}$

$Q_{f}$ = (1.63) (3.33 x 10⁵)

$Q_{f}$ = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q = Amount of energy remaining to raise the temperature

Using conservation of energy

E = $Q_{f}$ + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

T = 45 °C

5 0

2 years ago