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3 years ago

Is it true the atoms in sulphur liquid are the same as the atoms in solid sulphur?

1 answer:

3 0

Yes, it is true that the atoms in liquid sulphur are the same as the atoms
in solid sulphur.

Also, if you can heat it past 832°F, it'll boil, and you'll have sulphur vapor.
Those atoms are also the same.

In any one of its three physical phases, you can also spell it as "sulfur"
without changing the nature of its atoms.

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Where does the energy that is used to ride a bicycle up a hill come from and how is it classified?

MariettaO [177]

Well when you eat you get energy and then you use it to ride up the hill sorry I don't know how it is classified

7 0

3 years ago

Read 2 more answers

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

Is ‘velocity’the displacement of an object during specific unit of time?

babunello [35]

Answer:

Yes

Explanation:

The displacement covered by a body in unit time is called velocity.

5 0

3 years ago

Read 2 more answers

Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

STatiana [176]

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration= $\frac{v-u}{t}$

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

deceleration=a= $-9m/s^2$

We know that

Force =ma

Using the formula and suppose mass of my body=m=40 kg

The force exerted by wall on you

Force= $40\times (9)=360N$

3 0

2 years ago

A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg

motikmotik

Answer:

$\omega_{f}=1.634\ rad/s$

Explanation:

given,

diameter of merry - go - round = 2.40 m

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg

initial angular momentum of the system

$L_i = I\omega_i$

$L_i =356\times 1.80$

$L_i =640.8\ kg.m^2/s$

final angular momentum of the system

$L_f = (I_{disk}+mR^2)\omega_{f}$

$L_f = (356 + 25\times 1.2^2)\omega_{f}$

$L_f= (392)\omega_{f}$

from conservation of angular momentum

$L_i = L_f$

$640.8= (392)\omega_{f}$

$\omega_{f}=1.634\ rad/s$

8 0

3 years ago

Read 2 more answers