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Juli2301 [7.4K]
2 years ago
6

An athlete with mass m running at speed v grabs a light rope that hangs from a ceiling of height H and swings to a maximum heigh

t of h1. In another room with a lower ceiling of height H/2, a second athlete with mass 2m running at the same speed v grabs a light rope hanging from the ceiling and swings to a maximum height of h2. How does the maximum height reached by the 2 athletes compare and why? (Pick one) a) The first athlete reaches a greater height, because the athlete swings on a longer rope; b) The second athlete reaches a greater height, because the athlete has a greater mass; c) The 2 athletes reach the same height, because the effect of the rope length offsets the effect of the athletes’ masses; d) The 2 athletes reach the same height, because the athletes run with the same speed.
Physics
1 answer:
-Dominant- [34]2 years ago
7 0

Answer:

d) The 2 athletes reach the same height, because the athletes run with the same speed.

Explanation:

In the whole process , kinetic energy is converted into potential energy .

1/2 m v² = mgh

v² = 2gh

h = v² / 2g

In this expression we see that height attained does not depend upon mass of the object . At the same time it also makes it clear that it depends upon velocity . As the velocity in both the cases are same , height attained by both of them will be same. Hence option d ) is correct.

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marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
2 years ago
A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t
Lena [83]

Answer:

  E =  k Q    1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

              dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

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              dq = λ dx

We substitute in the equation

           ∫ dE = k ∫ λ dx / x²

We integrate

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We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

           E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

           E = k λ  (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

             E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

             E =  k Q    1 / (x₀-x₂) (x₀-x₁)

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5 0
2 years ago
Read 2 more answers
A 1.0 104 kg spacecraft is traveling through space with a speed of 1200 m/s relative to Earth. A thruster fires for 2.0 min, exe
aniked [119]
We are given information:
m=1.0* 10^{4} kg \\ v=1200m/s \\ t=2min=120s \\ F = 25kN = 25000N

If we apply Newton's second law we can calculate acceleration:
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Now we can use this information to calculate change of speed.
a = v / t
v = a * t
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Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
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Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
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p = 9 *10^6 kg*m/s

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Answer:

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now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

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now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

\frac{B_1}{B_2} = \frac{L_2}{L_1}

now plug in all values in it

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B_2 = 1.71 mT

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