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Katyanochek1 [597]
2 years ago
8

A supersonic jet, with a mass of 21,000 kg, departs from its home airbase with a velocity of 400 m/s due east. What is the jet’s

momentum?
Answer is - 21,000 x 400m/s = 84,00000 m/s
Physics
1 answer:
disa [49]2 years ago
6 0

Answer:

Momentum  P is 840000kgm/s or 8.4 × 10^6

Explanation:

Data :

        Mass = 21000 kg

        Velocity = 400 m/s

So momentum is given as

      P = mv

      P = 21000×400

    P = 8400000 kgm/s

       P = 8.4 × 10^6

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The bicyclist accelerates with magnitude <em>a</em> such that

25.0 m = 1/2 <em>a</em> (4.90 s)²

Solve for <em>a</em> :

<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²

Then her final speed is <em>v</em> such that

<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)

Solve for <em>v</em> :

<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s

Convert to mph. If you know that 1 m ≈ 3.28 ft, then

(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h

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A 1-pound ball and a 100-pound ball are dropped from a height of 10 feet at the same time. in the absence of air resistance, ___
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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 581 km from the center of asteroid
mariarad [96]

Answer:

0.05806

Explanation:

m_x = Mass of asteroid x

m_y = Mass of asteroid y

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r_y = Distance from asteroid y = 581 km

m = Mass of asteroid

Force of gravity between asteroid x and the astronaut

F_1=\frac{Gm_xm}{r_x^2}\\\Rightarrow F_1=\frac{Gm_xm}{140^2}

Force of gravity between asteroid x and the astronaut

F_2=\frac{Gm_ym}{r_y^2}\\\Rightarrow F_2=\frac{Gm_ym}{581^2}

Here these two forces are equal as they are in equilibrium

\frac{Gm_xm}{140^2}=\frac{Gm_ym}{581^2}\\\Rightarrow \frac{m_x}{140^2}=\frac{m_y}{581^2}\\\Rightarrow \frac{m_x}{m_y}=\frac{140^2}{581^2}\\\Rightarrow \frac{m_x}{m_y}=0.05806

The ratio of the masses of the asteroid is 0.05806

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