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3 years ago

The total resistance of a circuit is 25. The voltage drop across the battery is 6.0v. What is the current

Physics

2 answers:

dem82 [27]3 years ago

8 0

I hope this helps :)

Alja [10]3 years ago

6 0

Using Ohm's Law V = IR
6.0v / 25 = 0.24 A
I = 0.24A

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An object is moving with a constant velocity of 278 m/s. How long will it take it to travel 7500 m, using the formula Delta X=Vt

rusak2 [61]

If the velocity is constant then the acceleration of the object is zero.
$a=0 (m/s^2)$
Thus when we apply the equation
$\Delta X=vt+(at^2/2)$
It remains
$\Delta X =vt$
or equivalent
$t=(\Delta X/v) =7500/278 =26.98 (seconds)$

7 0

3 years ago

What is the process called in which water vapor and carbon dioxide retain heat

antoniya [11.8K]

The process in which water vapor and carbon dioxide retain heat is called the greenhouse effect.

<span>There are many gases which causes greenhouse effect like carbon dioxide, methane, nitrous oxide, fluorinated gases etc. These gases trap the heat in the atmosphere, resulting the surface of earth to get warm producing greenhouse effect.</span>

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3 years ago

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

masya89 [10]

The velocity would be the same for all ramps.

5 0

3 years ago

Read 2 more answers

Help pleasee

agasfer [191]

Answer:

i/f = i/o + i/i f = focal, o = object, i = image

1 / i = 1 / f - 1 / o = (o - f) / o f

i = o * f / ( o - f) image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

M = - i / o = = 28.9 / 12.5 = 2.3 magnification (convex lens)

8 0

3 years ago

A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

mina [271]

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

6 0

3 years ago