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2 years ago

14

Please help me with this question. BRAINLIEST if answered.

1 answer:

IceJOKER [234]2 years ago

8 0

... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.

... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
That's (1/30 + 1/10 + 1/15)⁻¹ = 5 ohms.

... The 5-ohm-equivalent block and the 20-ohm resistor form a
voltage divider across the battery.
The voltage across the 5-ohm-equivalent block is (5/25 x 30v) = 6v .

... The top branch of the block is equivalent to a (9 + 6) = 15-ohmer.
With 6v across its ends, the current through that branch is (6/15) = 0.4A .

... With 0.4A flowing through it, the 9-ohm resistor is dissipating

I²R = (0.4A)² (9 ohms) = (0.16 A²) (9 ohms) = 1.44 W (choice-3)

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Calcular la energía cinética de un cometa cuya masa es de 5×10 elevado a 31 kg y se mueve con velocidad de 216000km/h

PolarNik [594]

The kinetic energy is $9\cdot 10^{40}J$ .

Explanation:

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

K is the kinetic energy of the object

m is the mass of the object

v is the speed of the object

For the comet in this problem, we have:

$m=5\cdot 10^{31} kg$ is its mass

$v=216,000 km/h$ is the speed

First, we convert the speed from km/h to m/s:

$v=216,000 \frac{km}{h} \cdot \frac{1000 m/km}{3600 s/h}=60,000 m/s$

Therefore, the kinetic energy of the comet is

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2 years ago

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i

Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

$y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]$ (1)

n = 4

Now,

We know the standard eqn is given by:

$y = AsinKxcos\omega t$ (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = $14.4 m^{- 1}$

$\omega = 166\ rad/s$

where

A = Amplitude

K = Propagation constant

$\omega$ = angular velocity

Now, to calculate the string's wavelength,

(a) $K = \frac{2\pi}{\lambda}$

where

K = propagation vector

$\lambda = \frac{2\pi}{K}$

$\lambda = \frac{2\pi}{14.4} = 0.436\ m$

(b) The length of the string is given by:

$l = \frac{n\lambda}{2}$

$l = \frac{4\times 0.436}{2} = 0.872\ m$

(c) Now, we first find the frequency of the wave:

$\omega = 2\pi f$

$f = \frac{\omega}{2\pi}$

$f = \frac{2\pi}{166} = 26.42\ Hz$

Now,

Speed of the wave is given by:

$v = f\lambda$

$v = 26.419\times 0.436 = 11.518\ m/s$

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2 years ago

NEED HELP NOW WILL GIVE BRAINLIEST

Arturiano [62]

Answer:

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Explanation:

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pls mark me brainliest its the right answer

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