If a 2 kg ball has 120 J of PE, then how high off the ground is it?

Suppose that the model presented by student 1 is correct. Based on the information provided, what would be the bond angle in a m

Ugo [173]

Question: Predicting the shape of a molecule is relatively straight forward. A molecule's shape will always be determined by the number of electron pairs around the central atom. The number of electron pair corresponds to the number of atoms that are bound to the central atom of the molecule. For example, water contains two hydrogen atom bound to one atom of oxygen, giving the molecule a linear geometry.

Suppose that the model presented by student 1 is correct. Based on the information provided, what would be the bond angle in a molecule of perchlorate ion.

Answer: Suppose that the model presented by student 1 is correct The (perchlorate ion) will be a tetrahedral shape, O-Cl-O bond angle 109.5 due to four groups of bonding electrons and no lone pairs of electrons.

8 0

3 years ago

Two uniform solid spheres, A and B have the same mass. The radius of sphere B is twice that of sphere A. The axis of rotation pa

Svetach [21]

Answer:

Sphere B has 4 times more inertia than sphere A.

Explanation:

Inertia on a Solid sphere is given by:

$I = 2/5*M*R^2$

For this problem:

ma = mb = M

Rb = 2*Ra

With these values:

$Ia = 2/5*M*Ra^2$

$Ib = 2/5*M*(2*Ra)^2 = 4* (2/5*M*Ra^2)$

As you can see, Ib = 4 * Ia.

4 0

3 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

$m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)$

where:

$m_g = 3.0 kg$ is the mass of the gold bar

$C_g=129 J/kg C$ is the specific heat of gold

$T_g=99 C$ is the initial temperature of the gold bar

$m_w = 0.22 kg$ is the mass of the water

$C_w=4186 J/kg C$ is the specific heat of water

$T_w=25 C$ is the initial temperature of the water

$T_f$ is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

$m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C$

5 0

3 years ago

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

3 years ago

When a car suddenly stops at a red light, a book lying on the car seat slides forward. Why does the book continue to move forwar

goblinko [34]

Answer:

C

Explanation:

The answer is c because Newton’s third law states “ for every action there is an equal and opposite reaction.” Since the car quit moving the book did not stay at rest because there was a reaction.

3 0

3 years ago

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