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Kamila [148]
3 years ago
5

You are at the top of the Empire State Building on the 102nd floor, which is located 373 m above the ground, when your favorite

superhero flies over the building parallel to the ground at 50.0 % the speed of light. You have never seen your favorite superhero in real life. Out of curiosity you calculate her height to be 1.73 m. If the superhero landed next to you, how tall would she be when standing?
Physics
1 answer:
Liula [17]3 years ago
3 0

Answer:

1.5 m

Explanation:

H = actual height of the superhero = ?

H₀ = height of the superhero as observed = 1.73 m

v = speed of the superhero = 0.50 c

Using the equation

H = H_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}

Inserting the values

H = 1.73 \sqrt{1 - \left ( \frac{0.50 c}{c} \right )^{2}}

H = 1.5 m

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In the diagram, the liquid is vaporizing at which point?
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E is the vapourising state
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A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T
Zigmanuir [339]

The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :

<em>PE</em> = <em>m g h</em>

<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)

<em>PE</em> = 58.8 J

Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :

<em>PE</em> = <em>KE</em>

58.8 J = 1/2 <em>m v</em> ²

117.6 J = (2.0 kg) <em>v</em> ²

<em>v</em> = √((117.6 J) / (2.0 kg))

<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s

3 0
3 years ago
An object is attached to a vertical spring and bobs up and down between points a and
Anton [14]
 it would be at either A or B. 
7 0
2 years ago
What force does a trampoline have to apply to a gymnast to accelerate her straight up at ? Note that the answer is independent o
Andrew [12]

Answer: Force applied by trampoline = 778.5 N

<em>Note: The question is incomplete.</em>

<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary. </em>

Explanation:

The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.

= (m * a) + (m * g)

= m ( a + g)

= 45 kg ( 7.50 *  9.80) m/s²

Force applied by trampoline = 778.5 N

5 0
2 years ago
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what
Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

E=\dfrac{k}{d^2}

We can write it as follows :

\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2

Put all the values,

\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}

So, the correct option is (a).

5 0
2 years ago
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