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3 years ago

What is the potential difference across a 15 Ω resistor that has a current of 3.0 A?

Physics

2 answers:

Zepler [3.9K]3 years ago

7 0

Answer:

45V

Explanation:

R=V/I

V=I/R

V=15ohms•3.0 A

V=45V

ehidna [41]3 years ago

5 0

Answer:

45 V

Explanation:

Use Ohm's law:

V = IR

V = (3.0 A) (15 Ω)

V = 45 V

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How will you charge a metal sphere by the process of induction?

zzz [600]

<span>A Conducting Sphere Is On Top Of An Insulating Stand. So A Negatively Charged Tube Is Brought Near The Neutral Sphere Then It Forces Electron Movement From The Left To The Right Side Of The Sphere. Once Touched By The Ground, The Electrons Leave The Sphere. When The Tube Is Moved Away, There's An Overall Positive Charge Left On The Sphere.</span>

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3 years ago

What is Angular acceleration, please explain this concept. Give any equations that involve angular acceleration and explain them

Eva8 [605]

Answer:

Angular acceleration is defined as the rate of change of angular velocity of a body.

consider the attached figure as shown

It rotates with an angular velocity $\omega$

An point inside the object rotates along the path as indicated thus turning by an angle $\theta$ in time 't'

Thus we have

$\alpha =\frac{d\omega }{dt}\\\\=\frac{d}{dt}(\frac{d\theta }{dt})\\\\\therefore \alpha =\frac{d^{2}\theta }{dt^{2}}$

physically angular acceleration can be understood as the rate at which the angular speed of any object is changing with time.

8 0

3 years ago

A skier goes down a slope and detaches from the ground moving in the horizontal direction with a speed of 25m / s. The slope has

zepelin [54]

Answer:

107 m down the incline

Explanation:

Given:

v₀ₓ = 25 m/s

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

-Δy/Δx = tan 35°

Find: d

First, find Δy and Δx in terms of t.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (0 m/s) t + ½ (-10 m/s²) t²

Δy = -5t²

Δx = v₀ₓ t + ½ aₓ t²

Δx = (25 m/s) t + ½ (0 m/s²) t²

Δx = 25t

Substitute:

-(-5t²) / (25t) = tan 35°

t/5 = tan 35°

t = 5 tan 35°

t ≈ 3.50 s

Now find Δy and Δx.

Δy ≈ -61.3 m

Δx ≈ 87.5 m

Therefore, the distance down the incline is:

d = √(x² + y²)

d ≈ 107 m

8 0

3 years ago

Which describes how scientific models are used? (2 points) They allow scientists to make predictions. They prevent scientists fr

luda_lava [24]

I think the correct answer would be the first option. Scientific models are used to allow scientists to extrapolate or interpolate data. Or in other words, allow them to predict values without performing more experiments. However, they are obtained by experiments first.

4 0

3 years ago

3. Alpha Centauri A and B are Sun-like stars, and together they form the binary star Alpha Centauri AB. Alpha Centauri A has 1.1

IgorLugansk [536]

Answer:

8722.8 m/s, average speed of B relative to A (approximate)

Explanation:

The total mass of the two stars,

M = M₁ + M₂ = 2.007 M☉ = 3.99101985e+30 kg

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

GM = 2.663726378e+20 m³ sec⁻²

The orbital period,

P = 79.91 years = 2.521767816e+9 sec

The semimajor axis of the orbit,

a = ∛[P²GM/(4π²)]

a = 3.50090e+12 meters

Assuming that the orbit is circular, with radius equal to the calculated semimajor axis, the average speed in orbit,

V = C/P

where

C = 2πa = 2.19968e+13 meters

V ≈ 8722.8 m/s, average speed of B relative to A (approximate)

However, we can look up the eccentricity of the orbit of α Cen A relative to α Cen B, and then we can calculate the average speed in orbit more accurately.

e = 0.5179

C = 4a ∫(0,π/2) √(1−e²sin²θ) dθ

C = 2.04379e+13 meters

V = 8104.6 m/s, average speed of B relative to A (more accurate)

As you can see, the more nearly correct value is considerably different than the approximation obtained when the orbit is assumed circular.

Incidentally, we can find that

The periapsis distance is 1.68779e+12 meters.

The apoapsis distance is 5.31402e+12 meters.

The semilatus rectum is 2.56189e+12 meters.

The semiminor axis is 2.99482e+12 meters.

The focal parameter is 4.94669.

We can also calculate the separation of A and B when the orbital speed has this average value:

r = 2a { 1 + [ (2/π) ∫(0,π/2) √(1−e²sin²θ) dθ ]² }⁻¹

r = 3.75778349e+12 meters

Moreover, that separation occurs when the stars reach the part of their orbit having true anomalies of

θ = ± arccos{ [ 1 − (a/r) (1+e²) ] / e }

θ = 110.518° and 249.482°

3 0

3 years ago