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Paladinen [302]

3 years ago

7

Una ave vuela a una velocidad constante de 15m/s en una trayectoria rectilínea. Si dura una hora volando ¿cuanta distancia habrá

recorrido?

1 answer:

earnstyle [38]3 years ago

3 0

Answer:

54,000

Explanation:

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An ice cream truck is going 25m/s to East. It accelerates to 40m/s in the same direction over 3s. What is its acceleration

skad [1K]

Its acceleration is 5m/s2.
The formula for acceleration is Final speed - initial speed/time.
So 40m/s - 25m/s is 15m/s.
15m/s divided by 3s is 5m/s2.
Thus the acceleration is 5m/s2

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Answer:

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Explanation:

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Imagine that the light you observe from space is exhibiting a redshift. This would mean that the Universe is Choose one: A. trav

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Answer: C. Expanding

Explanation:

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3 years ago

What unit is used to measure<br> wavelength?

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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

<u>Now the net force</u>

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

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3 years ago