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Rasek [7]
3 years ago
10

Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one

is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Physics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

e.

Explanation:

  • Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according  Newton's 2nd Law, as we can see below:
  • F_{g}  = m*g = m*a  (1)
  • ⇒a = g = 9.8m/s² (pointing downward)
  • Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:

       \Delta y = v_{o} * t - \frac{1}{2} *g*t^{2}  (2)

  • Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:

        h =\frac{1}{2} *g*t^{2}  (3)

  • Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.
  • If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.
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Answer:

D

Explanation:

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3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
Two forces that are not equal in size are
Bumek [7]
They are unbalanced forces ..... Hope this helps :3
4 0
3 years ago
If the force on a hammer is 24 N and its mass is 1.6 kg, then the
Gnesinka [82]

Answer:

15 m/s²

Explanation:

F = ma

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a = F/m

6 0
3 years ago
A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

v=\omega r

where \omega is the angular speed and r is the radius of the circular path.

\omega is given by

\omega = 2\pi f

where f is the frequency of revolution.

Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

6 0
3 years ago
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