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kodGreya [7K]
3 years ago
10

Convert 8765 pounds to tons using one step conversion please help me heheeehehehhee

Physics
2 answers:
Andre45 [30]3 years ago
5 0

Answer:

4.3825 tons

Explanation:

Hope it helps :)

const2013 [10]3 years ago
4 0

The one-step conversion you need is (1 ton / 2000 pounds).

It's a fraction that's equal to ' 1 ', because the top and the bottom are equal.

Multiply the 8,765 pounds by it.  That won't change the weight.  It'll just change the units.

(8,765 pounds) x (1 ton / 2000 pounds) = <em>4.3825 ton</em>

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a 1600 kg car on flat ground is moving 6.25 m/s. its engine creates 1150 N forward force as the car moves 45.8 m. what is it fin
Sveta_85 [38]

Answer:

83,900 J

Explanation:

First, find the acceleration:

F = ma

1150 N = (1600 kg) a

a = 0.719 m/s²

Now find the final velocity.

Given:

Δx = 45.8 m

v₀ = 6.25 m/s

a = 0.719 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (6.25 m/s)² + 2 (0.719 m/s²) (45.8 m)

v = 10.2 m/s

Now find the final KE:

KE = ½ mv²

KE = ½ (1600 kg) (10.2 m/s)²

KE = 83,920 J

Rounded to three significant figures, the final kinetic energy is 83,900 J.

6 0
4 years ago
What do the arrows represent? gases mixing with particles liquid moving through a vacuum moving particles electromagnetic waves
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3 years ago
A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

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Explanation:

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Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

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4 years ago
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Igoryamba

Answer:

Increases, increases

Explanation:

The current is directly proportional to the voltage and inversely proportional to the resistance. The implication of this is that, whenever the voltage is increased, the current increases simultaneously. On the other hand, if the resistance is increased, the current will decrease accordingly and vice versa.

Recall that power is given by P= V^2/R where;

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We can see that power and resistance are inversely related hence decreasing the resistance increases the power output of the lightbulb.

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