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3 years ago

What line in the sky is created by our revolution around the Sun?

Physics

1 answer:

faust18 [17]3 years ago

6 0

Answer:

Ecliptic

Explanation:

You asked for the term that is used to describe Earth going around the sun

The ecliptic is the path the sun, moon, and planets take across the sky as seen from Earth. It best defines the plane of the Earth's orbit around the sun. The imaginary line can best be visualized in the days just before full moon.

Hope it helped

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5) (64(6-2) = A) 62 B) 8-9 C) 122 12-8

skelet666 [1.2K]

Answer:

B 8-9 because you had to subtract that number or simplify. then your answer is 8-9

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3 years ago

What is the force applied to a baseball that has a mass of 142kg and has a acceleration of 30m/s to the power of 2

aev [14]

Mass=m=142kg
Acceleration=a=30m/s
Force=F

Using Newton's second law

$\\ \sf\Rrightarrow F=ma$

$\\ \sf\Rrightarrow F=142(30)$

$\\ \sf\Rrightarrow F=4260N$

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3 years ago

Give 2 reasons for fitting heavy commercial vehicles with many tyres

forsale [732]

Better weight distribution and more stability

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3 years ago

a seismic wave has an amplitude of 0.012 Meters.If the amplitude of this wave reduces to 0.006 meters, what happens to the energ

irina1246 [14]

Answer:The energy of the wave by a factor of 4

Explanation:

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4 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$