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2 years ago

A bridge that was 5m long has been washed out by a heavy rain. The road on the

Physics

1 answer:

amid [387]2 years ago

8 0

Hello so first of all you put 2m in the and you timesand then i’ve got your answer

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HIIIIIIIIIIIIIIIIIII WHAT IS 50X550

jekas [21]

50 x 550 is 27500!!!!

3 0

3 years ago

14 kilometers per hour to meters per second

Alina [70]

Answer:

About 3.88 meters per second

Explanation:

14km= 14,000m

14,000m per hour

so 14,000 ÷ 60= 233.33 meters per minute

233.33÷60=3.88 meters per second

3 0

3 years ago

At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev

olganol [36]

Answer:

the aircraft must travel at a speed of 73.4 m/s in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

v₂ = 73.4 m/s

Therefore, the aircraft must travel at a speed of 73.4 m/s in order to create the ideal lift.

5 0

3 years ago

How is light described?

Lelu [443]

Light as a wave: Light can be described (modeled) as an electromagnetic wave. ... This changing magnetic field then creates a changing electric field and BOOM - you have light. Unlike many other waves (sound, water waves, waves in a football stadium), light does not need a medium to “wave” in.

Explanation:

5 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago