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inn [45]
3 years ago
8

What is a non contact force that attracts all objects to the centre of the earth

Physics
1 answer:
beks73 [17]3 years ago
4 0

Answer:

<em>Gravity</em><em>.</em><em> </em><em>The</em><em> </em><em>weight-force</em><em> </em><em>or</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>because</em><em> </em><em>of</em><em> </em><em>Gravity</em><em>,</em><em> </em><em>which</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>object</em><em> </em><em>attracting</em><em> </em><em>it</em><em> </em><em>towards</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>earth</em><em>.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>

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A steel beam that is 6.50 m long weighs 336 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
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Answer:

Explanation:

When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then

R₁ +R₂ = 336 + 590

= 929

Now the moment beam begins to tip , reaction on distant support R₁ = 0

only R₂ will exists on the support near to Suki.

Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.

336 x 1.5 = 590 x

x = .85 m

ie , from second support , Suki can not go beyond a distance of .85 m towards the second end.

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An 89.0 kg fullback moving east with a speed of 5.6 m/s is tackled by an 85.0 kg opponent running west at 2.84 m/s, and the coll
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Answer:

a. v_f=1.477m/s

b. ΔK=1558.3J

c. E_k=1034.7 J

Explanation:

a).

Momentum conserved

p_{ix}=p_{fx}

m_1*v_1+m_2*v_2=(m_1+m_2)*v_f

v_f=\frac{m_1*v_1+m_2*v_2}{m_1+m_2}

v_f=\frac{89.0kg*5.6m/s+85.0kg*-2.84m/s}{(89.0+85.0)kg}

v_f=1.477m/s

b).

ΔK=K_i-K_f

\frac{1}{2}*m_1*v_1^2+\frac{1}{2}*m_2*v_2^2=\frac{1}{2}*(m_1+m_2)*v_f^2

\frac{1}{2}*89.0kg*(5.6m/s)^2+\frac{1}{2}*85.0kg*(2.84m/s)^2=\frac{1}{2}*(89.0+85.0)kg*(1.447m/s)^2

ΔK=1558.3J

c).

E_k=\frac{1}{2}*89kg*(5.8m/s)^2-\frac{1}{2}*(85+89)kg*(1.44m/s)^2

E_k=1034.7 J

d).

All of which has been lost as mechanical energy, and is now thermal energy warmer football players, noise a loud crunch for example.

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Pwease hlp me (&gt;O&lt;) Darkness falls during daytime hours. Which event might have occurred?
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Answer:

D. Solar Eclipse

Explanation:

This is quite a phenomenal occurrence. Please, allow me to explain why D is correct.

A. A New Moon: This is the moon's first lunar phase. This is such a phase during which the sun and earth are located at nearly precise opposite ends of the moon, an occurrence that renders the moon nearly (if not) entirely invisible by the naked eye from an earthly perspective. This is not an occurrence that renders darkness during daylight.

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D. A Solar Eclipse: A solar eclipse is a breathtaking and oftentimes somewhat terrifying occurrence during which the moon passes <em>perfectly </em>in between the earth and the sun. Unlike a New Moon, however, this occurrence is perceived <em>during the day</em>, so for the side of the earth FACING the sun. When the moon is making its journey in between the sun and earth (which takes roughly 3-5m), the massive silhouette of the moon can be seen blotting out the sun, effectively causing a brief moment of darkness during daytime hours. This occurence is considered remarkably rare because DESPITE it occuring once per year, it can ONLY be viewed from specific locations on earth.

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