Menciona un ejemplo de la vida cotidiana donde se apliquen las cuatro fuerzas fundamentales de la naturaleza.

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

or

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

or

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

or

I = $0.0109[1-2.717^{-1}]$

or

I = 0.00688 A

or

I = 6.88 mA

5 0

3 years ago

An injured monkey sits perched on a tree branch 3.0 m above the ground, while a wildlife veterinarian is kneeling down in the bu

Irina18 [472]

here in the given situation if monkey starts free fall at the same instant when veterinarian shoots towards it then we know that vertical component of motion of monkey and the dart will be same as under gravity

so here the dart will always hit the monkey because they both moves under same acceleration

so here for the angle we can use

$tan\theta = \frac{H}{L}$

now we have

H = 3 m

L = 87.5 m

now we will have

$tan\theta = \frac{3}{87.5}$

$tan\theta = 0.034$

$\theta = 1.96 degree$

so angle will be 1.96 degree above the ground

6 0

3 years ago

A 4000 kg rocket is launched, shooting 50 kg of burned fuel from its exhaust at a velocity of -625 m/s. What is the velocity of

AnnZ [28]

Hopefully this will help you.

8 0

4 years ago

43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a

olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

$P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J$

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J$

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

$\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N$