I really need help with this assignment

A child picks up a tennis ball tied to the end of a rope. He swings the rope around, over his head, in a circle. Once the ball i

kkurt [141]

Answer:

The ball will fly tangential to the original circle

Explanation:

The image here is missing, however we can still answer to the question.

In fact, the circular motion of the ball when it is tied to the rope is a combination of two separate effects:

1- The centripetal force, in the form of the tension in the rop, that pulls the ball at any time towards the centre of the circular path

2- The inertia of the ball, which tends to continue its motion in a straight direction, tangential to the circle and perpendicular to the direction of the centripetal force

When child let the string go, there is no more tension in the string acting on the ball, and therefore, there is no longer a centripetal force.

As a result, number 1) disappears, and therefore there is only the inertia of the ball that will determine its motion: and therefore, the ball will continue its motion straight in a direction tangential to the original circle.

5 0

3 years ago

Which statement describes a chemical property of an object? A:The object is white in color.B:The object has a powdery texture.C:

kozerog [31]

Answer:

D

Explanation:

Color, texture, and density are all physical properties but reactivity is a chemical property so the answer is D.

3 0

3 years ago

Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling

Neko [114]

Answer:

104.352°C

Explanation:

Data Given:

Boiling point of water = 100.0°C

Kb (boiling point constant = 0.512°C/m

Concentration of the Mg₃(PO₄)₂ = 8.5 m

Solution:

Formula Used to find out boiling point

ΔTb = m.Kb . . . . . . (1)

where

ΔTb = boiling point of solution - boiling point of water

So,

we can write equation 1 as under

ΔTb = Tb (Solution) -Tb (water)

As we have to find out boiling point so rearrange the above equation

Tb (Solution) = m.Kb + Tb (water) . . . . . . . (2)

Put values in Equation 2

Tb (Solution) = (8.5 m x 0.512°C/m ) + 100.0°C

Tb (Solution) = 4.352 + 100.0°C

Tb (Solution) = 104.352°C

so the boiling point of Mg₃(PO₄)₂ 8.5 m solution = 104.352°C

3 0

3 years ago

How many moles of O2 are required to react with 6.6 moles of H2?

Bingel [31]

Answer:

For 1: 3.3 moles of oxygen gas is required.

For 2: 14 moles of hydrogen gas is required.

For 3: 1.5 moles of oxygen gas is required.

Explanation:

The chemical reaction of oxygen and hydrogen to form water follows:

$O_2+2H_2\rightarrow 2H_2O$

For 1: When 6.6 moles of $H_2$ is reacted.

By Stoichiometry of the above reaction:

2 moles of hydrogen gas reacts with 1 mole of oxygen gas.

So, 6.6 moles of hydrogen gas will react with = $\frac{1}{2}\times 6.6=3.3mol$ of oxygen gas.

Hence, 3.3 moles of oxygen gas is required.

For 2: When 7.0 moles of $O_2$ is reacted.

By Stoichiometry of the above reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas.

So, 7 moles of oxygen gas will react with = $\frac{2}{1}\times 7=14mol$ of hydrogen gas.

Hence, 14 moles of hydrogen gas is required.

For 3: When 3.0 moles of $H_2O$ is formed.

By Stoichiometry of the above reaction:

2 moles of water is formed from 1 mole of oxygen gas.

So, 3.0 moles of water will be formed from = $\frac{1}{2}\times 3.0=1.5mol$ of oxygen gas.

Hence, 1.5 moles of oxygen gas is required.

7 0

3 years ago

At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO

erastovalidia [21]

Answer : The concentration of $NO$ at equilibrium is 0.9332 M

Solution : Given,

Concentration of $N_2$ and $O_2$ at equilibrium = 0.200 M

Concentration of $N_2$ and $O_2$ at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}$

$K_c=6.25$

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}$

By solving the term x, we get

$x=2.6\text{ and }-0.0666$

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of $NO$ at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of $NO$ at equilibrium is 0.9332 M

5 0

4 years ago