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4 years ago

Airborne objects tend to curve to the right

Physics

2 answers:

Alika [10]4 years ago

6 0

Airborne objects tend to turn right in the northern hemisphere, caused by the Coriolis force.

Paraphin [41]4 years ago

5 0

Answer:

in the northern hemisphere

Explanation:

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Si tu velocidad al caminar es 0,3 m/s. Determina el tiempo que demorarias en llegar al sol en horas (REALIZAR TRANSFORMACIÓN DE

Contact [7]

Answer:

138,516,546.9 horas.

Explanation:

Tenemos que usar la ecuación:

Velocidad = distancia/tiempo

Acá tenemos:

Velocidad = 0.3m/s

distancia = 149597870700 m

y queremos resolver la ecuación para el tiempo:

0.3m/s = 149597870700m/tiempo.

tiempo = 149597870700m/(0.3m/s) = 498,659,569,000 s

y sabemos que una hora tiene 3600 segundos, entonces si queremos transformar de segundos a horas tenemos:

498,659,569,000 s = (498,659,569,000/3600) h = 138,516,546.9 horas.

6 0

3 years ago

ASAP NEED HELP PLEASE

Sholpan [36]

Answer:

57.91

Explanation:

5 0

3 years ago

How much work is done when a hoist lifts a 290-kg rock to a height of 7 m?

Zepler [3.9K]

Given: Mass m = 290 Kg; Height h = 7 m

Required: Work = ?

Formula: Work = Force x distance but, Force F = mg

W= fd

W = mgh

W= (290 Kg)(9.8 m/s²)(7 m)

W = 19,894 Kg.m²/s²

W = 19,894 J

5 0

4 years ago

A 250-kN railroad car A is traveling at 40 m/s while a 550-kN freight car B is traveling at 10 m/s (both cars heading to the rig

Damm [24]

Answer:

the maximum deformation undergone by the spring = 47.46 cm

Explanation:

Using conservation of momentum:

$m_Av_A + m_Bv_B = (m_A+m_B )v$

where:

$m_A = \frac{250\ kN}{g}$

$v_A = 40$

$m_B = \frac{550\ kN}{g}$

$v_B =10$

Then;

$m_Av_A + m_Bv_B = (m_A+m_B )v$

$\frac{250*10^3}{9.81}*40 + \frac{550*10^3}{9.81}*10 = (\frac{800*10^3}{9.81} )v$

$1580020.387 = 81549.43935 \ v$

$v = \frac{1580020.387}{81549.43935}$

v = 19.375 m/s

However ; using conservation of energy to determine the maximum deformation undergone by the spring ; we have:

$\frac{1}{2} [m_Av_A^2 +m_Bv_B^2] =\frac{1}{2}[(m_A+m_B)v^2 + kx^2]$

$[m_Av_A^2 +m_Bv_B^2] =[(m_A+m_B)v^2 + kx^2]$

$[\frac{250*10^3}{9.81}*40^2 + \frac{550*10^3}{9.81}*10^2] =[ (\frac{800*10^3}{9.81} )*19.375^2 + 70 *10^6 \ * x^2]$

x = 0.4746 m

x = 47.46 cm

Thus, the maximum deformation undergone by the spring = 47.46 cm

3 0

4 years ago

Read 2 more answers

Can someone help me please

Svetllana [295]

Answer:

B.) Visit a healthcare professional

Explanation:

Their judgement is much better than a friend's or your own.

6 0

3 years ago