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Karo-lina-s [1.5K]

3 years ago

9

Beryllium has a charge of 2, and bromine has a charge of –1. which is the best name for the ionic bond that forms between them?

beryllium bromide beryllium bromine bromine berylliumide bromide berylliumide

1 answer:

Otrada [13]3 years ago

8 0

The best name for the ionic bond that forms between them is Beryllium Bromide.

We have been provided with data,

Beryllium charge, q = 2

Bromine charge, q = -1

As we know the valance electron of Be is +2 and the valance electron of bromine is -1. Since one is metallic and the other is non-metallic.

Now, when they combine they exchange valance electron, and bromine change into bromide so they form Beryllium Bromide.

So, the best name for the ionic bond that forms between them is Beryllium Bromide.

Learn more about ionic bonds here:

brainly.com/question/21464719

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My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0 ⇒ K₀

and h₁ = 0 ⇒ U₁ = 0

we get

U₀ = K₁

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0

3 years ago

A skier is standing on a horizontal patch of snow. She is holding onto a horizontal tow rope, which is about to pull her forward

fredd [130]

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move will be 81.03 N,

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

It is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration. Mathematically aion in the different components and balancing the equation gets. Components in the x-direction.

The normal force is balanced by weight;

N = mg

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move is;

F = μN

F= μmg

F=0.14 × 59 kg × 9.81 m/s²

F = 81.03 N

Hence, the magnitude of the maximum force will be 81.03 N,

To learn more about the friction force refer to the link;

brainly.com/question/1714663

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5 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

4 years ago

(a) A standard sheet of paper measures 8 1/2 by 13 inches. Find the area of one such sheet of paper in m2.8.5(!meter/39.37in) —

Sergeu [11.5K]

Answer:

a) A = 0.07129 m²

b) A / A ’= 1.77

Explanation:

In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.

width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)

a = 0.2159 m

length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)

l = 0.3302 m

The area of a rectangle is

A = l a

A = 0.3302 0.2159

A = 0.07129 m²

b) we have a second sheet with reduced dimensions

a ’= 3/4 a

l ’= ¾ l

Let's find the area of this glossy sheet

A ’= l’ a ’

A ’= ¾ l ¾ a

A ’= 9/16 l a

To find the factor we divide the two quantities

A / A ’= l a 16 / (9 l a

A / A ’= 1.77

8 0

3 years ago

A scuba tank, when fully submerged, displaces 16.7 L of seawater. The tank itself has a mass of 13.4 kg and, when "full," contai

meriva

Answer:

$m = 17.87 kg$

Explanation:

As we know that tank itself has mass given as 13.4 kg

so it is given as

$m_1 = 13.4 kg$

also the tank contains air in it and the mass of air inside the tank is also given as 4.47 kg

so it is

$m_2 = 4.47 kg$

so total mass of the tank is the mass of the empty tank and mass of air in the tank

$m = m_1 + m_2$

$m = 13.4 + 4.47$

$m = 17.87 kg$

3 0

3 years ago