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3 years ago

Did the deadliest hurricanes happen recently or in the distant past

Physics

1 answer:

myrzilka [38]3 years ago

6 0

Answer:

Galveston hurricane of 1900

Explanation:

also called Great Galveston hurricane, hurricane (tropical cyclone) of September 1900, one of the deadliest natural disasters in U.S. history, claiming more than 8,000 lives.

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Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

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B. kinetic energy increases and potential energy decreases

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Latitude where each of the following types of winds-

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westerlies

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3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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3 years ago

I need help with science please, ill give brianlist’s :

Elis [28]

Answer:

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Explanation:

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