All categories

3 years ago

Explain the relationship between air resistance and gravitational acceleration on a falling object

1 answer:

7 0

Air resistance, also called drag, acts upon a falling body by slowing the body down to thr point where it stops accelerating, and it falls at a constant speed, known as the terminal volocity of a falling object. Air resistance depends on the cross sectional area of the object, which is why the effect of air resistance on a large flat surfaced object is much greater than on a small, streamlined object.

You might be interested in

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

The temperature inside my refrigerator is about 4 degrees C. If I place a balloon in my fridge that initially has a temperature

maksim [4K]

Convert the given temperatures from celsius to kelvin since we are dealing with gas.

To convert to kelvin, add 273.15 to the temperature in celsius.

T1 = 22 + 273.15 = 295.15 k

T2 = 4 + 273.15 = 277.15 k

V1 = 0.5 L

Let's find the final volume (V2).

To solve for V2 apply Charles Law formula below:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

5 0

1 year ago

A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

6 0

3 years ago

A falling raisin will have<br> whale. (more/less)<br> momentum than a falling blue

Natalka [10]

Answer:less

Explanation:thats the answer

3 0

2 years ago

URGENT!!

irinina [24]

Answer:

Explanation:La ecuación de Van der Waals es una ecuación de estado de un fluido compuesto de partículas con un tamaño no despreciable y con fuerzas intermoleculares, como las fuerzas de Van der Waals. La ecuación, cuyo origen se remonta a 1873, debe su nombre a Johannes van der Waals, quien recibió el premio Nobel en 1910 por su trabajo en la ecuación de estado para gases y líquidos, la cual está basada en una modificación de la ley de los gases ideales para que se aproxime de manera más precisa al comportamiento de los gases reales al tener en cuenta su tamaño no nulo y la atracción entre sus partículas.

3 0

3 years ago