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3 years ago

According to ohms law, as the voltage increases across a 40 ohm resistor what happens to the current, resistors, and resistance

Physics

1 answer:

irakobra [83]3 years ago

7 0

Answer:

As the voltage increases, the current flowing through the circuit increases while the resistance of the resistor remains constant.

Explanation:

Ohm's law states that the current flowing through a circuit is directly proportional to the applied voltage.

V = IR

where;

I is the current

R is the resistance

V is the applied voltage

Based on this law (Ohm's law), as the voltage increases, the current flowing through the circuit increases while the resistance of the resistor remains constant.

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The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

3 years ago

a car is traveling at velocity of 15m/s.it accelerates to a velocity 35m/s in 10s calculate the acceleration?

vlada-n [284]

Answer:

2m/s is acceleration I hope it helps you

3 0

2 years ago

Read 2 more answers

1.) if you are sitting still are you accelerating? How do you know?

castortr0y [4]

yes. gravity is working on you and everything on you.

also earth is rotating

no not in respect ofimmediate surroundings

object could move at constant velocity = no change in either magnitude or direction

change of position "proves" this

5 0

3 years ago

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.

Triss [41]

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

ρA₁V₁ = ρA₂V₂

A₁V₁ = A₂V₂

making V₂ the subject of the equation;

$V_{2} = \frac{A_{1}V_{1}}{A_{2}}$

the pipe is widened to twice its original radius,

r₂ = 2r₁

then the cross-sectional area A₂ = 4A₁

⇒ $V_{2}= \frac{A_{1}V_{1}}{4A_{1}}$

$V_{2}= \frac{V_{1}}{4}$

This implies that the water speed will drop by a factor of $\frac{1}{4}$ because of the increase the pipe cross-sectional area.

The Bernoulli Equation;

Energy per unit volume before = Energy per unit volume after

p₁ + $\frac{1}{2}$ ρV₁² + ρgh₁ = p₂ + $\frac{1}{2}$ ρV₂² + ρgh₂

Total pressure is constant and $P_{T}$ = P = $\frac{1}{2}$ ρV₂²ρV²

p₁ + $\frac{1}{2}$ ρV₁² = p₂ + $\frac{1}{2}$ ρV₂²

Making p₂ the subject of the equation above;

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρV₂²

But $V_{2} = \frac{V_{1}}{4}$ so,

p₂ = p₁ + $\frac{1}{2}$ ρV₁² - $\frac{1}{2}$ ρ $\frac{V_{1}^{2}}{4^{2}}$

p₂ = 3.0 x 10⁴ + ( $\frac{1}{2}$ × 1000 × 12²) - ( $\frac{1}{2}$ × 1000 × 12²/4² )

P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³

P₂ = 9.79 × 10⁴Pa

P₂ = 9.8 × 10⁴Pa

4 0

2 years ago

A car travels at a velocity of 8m/s for 16s . How far did it travel?

sertanlavr [38]

Answer:128m

Explanation:

8*16=128

7 0

3 years ago