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3 years ago

Will give brainliest!

1 answer:

8 0

Howdy!!

your answer is ----

total resistance 1/R = 1/9+1/9+1/9
= 3/9

=> R = 9/3 = 3 ohm
according to ohms law

voltage = current × resistance

= 4 × 3 = 12 volt

hope it help you

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

The missing energy has turned into interned energy in the completely inelastic collision

3 0

3 years ago

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

KatRina [158]

Answer:

The value is $x = 45.99 \ Au$

Explanation:

From the question we are told that

The period of the asteroid is $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

$R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value

$M = 1.99*10^{30} \ kg$

G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

$R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=> $R = 6.88 *10^{12} \ m$

Generally

$1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

$R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=> $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=> $x = 45.99 \ Au$

7 0

2 years ago

A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use

polet [3.4K]

The net force of the object is equal to the force applied minus the force of friction.
Fnet = ma = F - Ff
12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107.

7 0

2 years ago

How do the amplitudes of a 120 decibel sound and a 100-decibel sound compare?

NemiM [27]

The 120 decibel sound has more amplitude than the 100 decibel sound.

In Physics, the relation between amplitude and intensity is that the intensity of the wave is directly proportional to the square of its amplitude.

4 0

1 year ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$