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3 years ago

12

Like-charged bodies, when brought closer together, will: repel more strongly be neutralized repel less strongly attract more str

ongly

2 answers:

SVETLANKA909090 [29]3 years ago

5 0

Answer:

That's almost the true

Explanation:

it does not happen all the time

gogolik [260]3 years ago

3 0

The answer is A: Repel More Strongly

Explanation: I took the test but two alike magnets will push away form each other

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A man, facing a high wall, notices that an echo is heard 4.0 seconds after he makes a sharp sound. After walking 200m directly t

cestrela7 [59]

Answer:2.4

Explanation:basically 2.4 = 2.500 kg and then we add 3.4 seconds in all of them 4.0 + 3.4 = 2.5 ok

4 0

3 years ago

Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe

Mice21 [21]

As per the question the distance of venus from sun is given as 0.723 AU

We have been asked to calculate the time period of the planet venus.

As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically

$T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$ where is k is the proportionality constant

We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days

Hence $T_{1} =365.5 days$

The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun

Hence $R_{1} =1 AU$

The distance of venus from sun is 0.723 AU i.e $R_{2} =0.723$

From keplers law we know that- $\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{3} }$

⇒ $T_{2} ^{2} =T_{1} ^{2} *\frac{R_{2} ^{3} }{R_{1} ^{3} }$

Putting the values mentioned above we get-

$T_{2} ^{2} =50,350.132851075$

⇒ $T_{2} =\sqrt{50,350.132851075}$

⇒ $T_{2} = 224.388352752710 days.$

Hence the time period of venus is 224.388352752710 days

7 0

4 years ago

Read 2 more answers

What is the operating life time of a PV Module?

julia-pushkina [17]

Answer:

The industry standard life span is about 25 to 30 years, and that means that some panels installed at the early end of the current boom aren't long from being retired is the operating life time of a PV module .

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2 years ago

a thermometer has its steam marked in millimetre instead of degree celsius. The lower fixed points is 180 mm calculate the tempe

Nady [450]

Answer:hi

Explanation:

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3 years ago

A 165 pound football player assumed a 4 point stance at the line. If his base of support was 30” wide, 38” deep, with the center

Hatshy [7]

Answer:

(1) The front attack is = 1528.56 inch-pounds (2) From the side is = 1237.5 inch- pounds. (3) from the rear is = 1528.32 inch pounds

Explanation:

Solution

Given

The weight of a football player is = 65 pound

Instance = 4 point

Now,

Let us consider that the weight on both hands are the same and weight on both knees are also same.

Let say, weight on each hand be declared as m1, and weight of each leg be m2

Then,

m₁+ m₁ + m₂ +m₂ = 165 pounds

m₁ + m₂ = 82.5 ---------( equation 1)

On the coordinate plane let us assume that the center of gravity G is at the origin (0,0)

so,

2m₁x₁ + 2m₂x₂ /2m₁ + 2m₂ = 0

m₁ * 16 - m₂ * 22 = 0

m₂ = 8 /π m₁------ (2)

Now, let substitute 8 /π m for m₂ in equation 1

m₁ + 8 /π m₁ = 82.5

where m₁ = 47.76 pounds and m₂ = 34.74 pounds

Now,

(a) The front attack

The weight in front that is the one arm is displaced and about the center of gravity

so,

the inch of pound needed is denoted as:

Mfront = 2m₂ * x₂ = 2 * 34.74 *22

The Mfront becomes = 1528.56 inch-pounds

(2) From the side

The weight on one leg and one hand is displaced about the center of gravity G

Mside = 2/2 * y (m₁ +m₂) = 15¹¹ * (82.5)

so,

Mside = 15 * 82.5

Mside = 1237.5 inch- pounds

(c) For the rear attack

Now,

For the real attack, the weight in rear end on the knees is displaced about the center of gravity

Mrear = 2m₁ * x₂ = 2 * 47.76 * 16

Therefore the Mrear = 1528.32 inch pounds

4 0

3 years ago