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3 years ago

How does a 12-month lunar calendar differ from our 12-month solar calendar?.

Physics

1 answer:

horrorfan [7]3 years ago

8 0

Answer:

It has about 11 fewer days. It does not have seasons. Its new year always occurs in February instead of on January 1.

Explanation:

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A car with a momentum (impulse) of 20,000 kg m/s collides with a wall and comes to a rest in 0.1 seconds. How much

Pani-rosa [81]

Answer:

» Force is 200,000 Newtons

Explanation:

${ \tt{force = \frac{impulse}{time} }} \\ \\ { \tt{force = \frac{20000}{0.1} }} \\ \\ { \tt{force = 200000 \: newtons}}$

8 0

3 years ago

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More recent findings of children with exceptionally high IQs _

Alenkasestr [34]

Answer:

Explanation:

Researchers and educators differ in how they define gift-

edness. Traditionally, researchers have defined giftedness

as high general intelligence as measured by a high global

IQ score (Hollingwort

Exceptionally intelligent students (hereafter referred to

as gifted students) face a variety of problems in ordinary

classrooms. They often are ostracized as being different

and weird and are labeled as nerds and geeks (Silverman,

1993a, 1993b). In addition, they face the problem of

boredom due to lack of an appropriate level of challenge

(Csikszentmihalyi, Rathunde, & Whalen, 1993; Gross,

1993). Teachers often make little accommodation to the

needs of these children, and many teachers have little

or no special training in how to teach such exceptional

children

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3 years ago

Which is the work required to move a charge?

bulgar [2K]

Answer:

O electric potential

Explanation:

3 0

3 years ago

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A 1000 Kg car approaches an intersection traveling north at 30 m/s . A 1250 Kg car approaches the same intersection traveling ea

Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, $m_1=1000\ kg$

Initial velocity of first car is, $u_1=30\vec{j}\ m/s$

Mass of second car is, $m_2=1250\ kg$

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

$m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s$

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

$m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s$

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, $u_{2y}=0$ .

The magnitude of final velocity is given as:

$|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s$

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ$

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

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3 years ago

Which statements describe Rutherford model of the atom

il63 [147K]

Answer:

The nucleus is small and Very delicate. The electrons, for its turn it orbits around your nucleus. As the atom is the control/union of nucleus and electrons, and other particles, it is fair to say that an atom is a mostly empty space

Explanation:

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3 years ago

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