Ask question

Ask question

All categories

3 years ago

13

A pendulum is released from rest at point A. If the horizontal line represents the reference point, the pendulum has energy at p

oint B.

2 answers:

PSYCHO15rus [73]3 years ago

7 0

b i think

hope it helps ^^

Rzqust [24]3 years ago

6 0

Only kinetic

______________

Okay?

You might be interested in

As a result of nuclear fusion, the sun produces what 2 types of energy ??

ELEN [110]

Answer:

heat energy and nuclear energy

Explanation:

as a resukt of nuclear fusion, the sun produces two types of energy.

1. heat energy

2. nuclear energy

3 0

3 years ago

Of the three sciences, physics, chemistry, and biology, the most complex is

Agata [3.3K]

Answer:

I would probably say C to be completely honest

Explanation:

If you agree make sure to give me a like

6 0

2 years ago

¿Qué energía cinética posee una pelota de tenis de 65 g que se sirve a una velocidad de 200 km / h?

vekshin1

Therefore, the kinetic energy of an object is proportional to the square of its velocity (speed). In other words, If there is a twofold increase in speed, the kinetic energy will increase by a factor of four. If there is a threefold increase in speed, the kinetic energy will increase by a factor of nine.

4 0

3 years ago

Average velocity is different than average speed because calculating average velocity involves a)Distance b)Time c)Motion d)Disp

likoan [24]

Answer:

The answer is D.

Explanation:

Average speed involve just distance and time but average velocity includes displacement and time.

(Correct me if I am wrong)

6 0

4 years ago

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