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3 years ago

A solid ball is rolling along a horizontal surface at 3.7 m/s when it encounters an upward

Physics

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

h = 0.697 [m]

Explanation:

To solve this problem we must use the energy conservation theorem, where it tells us that kinetic energy is converted to potential energy or vice versa.

$E_{kinet}=E_{pot}$

where:

Ekinet = kinetic energy [J]

Epot = potential energy [J]

$\frac{1}{2}*m*v^{2}=m*g*h\\0.5*(3.7)^{2} =9.81*h\\h =(6.845)/9.81\\h = 0.697 [m]$

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A bike traveling initially at a speed of 32 m/s accelerates

Vika [28.1K]

We are given:

Initial velocity (u) = 32 m/s

Acceleration (a) = 3 m/s²

Displacement (s) = 40 m

Final Velocity (v) = v m/s

Solving for the Final Velocity:

from the third equation of motion:

v² - u² = 2as

replacing the variables

v² - (32)² = 2(3)(40)

v² = 240 + 1024

v² = 1264

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3 years ago

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Answer:

Explanation:

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3 years ago

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit

Reika [66]

The Answer is C Both at the same time

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3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

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To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

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3 years ago

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pishuonlain [190]

Answer & Explanation:

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3 years ago