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2 years ago

What is the total resistance in the circuit? (include unit in answer - ohms) I WILL GIVE BRAINLIEST!!

Physics

2 answers:

VashaNatasha [74]2 years ago

8 0

YES I TOOK THIS ! The answer is 25.79.

nadezda [96]2 years ago

3 0

Answer:

25.79

Explanation:

Give the person who answered before me BRAINLIEST, they deserve it.

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Calculate the momentum of a 1,500 kg car traveling at 6 m/s.

mamaluj [8]

Answer: 9000 kgm/s

Explanation:

Mass of car = 1500 kg

Speed by which car moves = 6 m/s. Momentum of the car = ?

Recall that:

Linear momentum = Mass x Speed

= 1500kg x 6m/s

= 9000 kgm/s

Thus, the linear momentum of the car is 9000 kgm/s

3 0

3 years ago

5. Why does a Ntion try to influence the value of its currency?

Mazyrski [523]

Answer:

If inflation is relatively lower than competitors, then the countries goods will become more attractive and demand will rise. Lower inflation tends to increase the value of the currency in the long term.

Explanation:

7 0

2 years ago

A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

Fantom [35]

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

- q × E = m × w² × r ..................1

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}r}$ q = m × w² × r ............2

and w = $\frac{2*\pi}{T}$

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$ × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ = 1.672 × $10^{-27}$ × 11304000² × 0.0120

linear charge density = -9.495 × $10^{-34}$ C/m

8 0

2 years ago

What does a lunar eclipse and a solar eclipse have in common

liraira [26]

During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".

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3 years ago

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)