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3 years ago

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Una gota de lluvia (m = 3.54 x 10-5 Kg.) cae verticalmente a velocidad constante bajo la influencia de la gravedad y la resisten

cia del aire. Después de que la gota ha descendido 121 m, ¿cuál es el trabajo (mJ) realizado por la gravedad?

1 answer:

konstantin123 [22]3 years ago

4 0

Answer:

Hola no speak Espanol

Explanation:

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Contrast elastic potential energy and chemical potential energy

tatuchka [14]

Chemical Potential Energy is released when chemical bonds between atoms are broken (like covalent and ionic) and is released mainly as thermal
<span>Elastic Potential is released when the molecules in the material are allowed to go back to there original form, and is released mainly as kinetic</span>

3 0

3 years ago

A curve that has a radius of 90 m is banked at an angle of =10.8∘. If a 1100 kg car navigates the curve at 75 km/h without skidd

PilotLPTM [1.2K]

The minimum coefficient of static friction between the pavement and the tires is 0.69.

The given parameters;

<em>radius of the curve, r = 90 m</em>
<em>angle of inclination, θ = 10.8⁰</em>
<em>speed of the car, v = 75 km/h = 20.83 m/s</em>
<em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

$F_n = mgcos(\theta)$

The frictional force between the car and the road is calculated as;

$F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)$

The net force on the car is calculated as follows;

$mgsin(\theta) + \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \ + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) + \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\$

$\mu_s = 0.69$

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

8 0

3 years ago

Someone answer these questions please???

chubhunter [2.5K]

Yeah sure someone else in answering rn

8 0

3 years ago

Read 2 more answers

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

4 0

3 years ago

What is the wavelenght and the frequencen of walkie-talkie?

GuDViN [60]

Nowadays, most of the walkie talkies ... and all R/C models ... operate in the "Family Radio Service" (FRS) band. That's 46-49 MHz. The wavelength is 6.12-6.52 meters.

8 0

3 years ago