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2 years ago

State the relative position for the sun and earth during the lunar eclipse (ASAP)

Physics

1 answer:

Mademuasel [1]2 years ago

8 0

Answer:

Solar eclipses result from the Moon blocking the Sun relative to the Earth; thus Earth, Moon and Sun all lie on a line. Lunar eclipses work the same way in a different order: Moon, Earth and Sun all on a line. In this case the Earth's shadow hides the Moon from view.

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Un auto de 2,300 kg está estacionado sobre una rampa inclinada 26.0°. Obtenga la tensión en la cadena.

Natasha_Volkova [10]

Answer:

T = mgsinθ = 2300(9.8)sin26.0 = 9880.88 ≈ 9900 N

Explanation:

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2 years ago

What is transmitted by all waves? a. energy b.mass c.matter b.sound

Evgesh-ka [11]

A. Energy is transmitted by all waves.

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2 years ago

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What is Gravity? PLEASE ANSWER

jolli1 [7]

Answer:

Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun.

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2 years ago

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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2 years ago

A 70.0 kg person climbs stairs, gaining 3.90 meters in height. Find the work done (in J) to accomplish this task.

Veronika [31]

Answer:

Work done, W = 2675.4 J

Given:

mass, m = 70.0 kg

height, H = 3.90 m

Solution:

According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:

Work done, W = mgH

where

g = acceleration due to gravity = $9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J$

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2 years ago