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KatRina [158]
3 years ago
14

. A bead of mass m under the influence of gravity slides along a frictionless and massless wire whose height is given by a funct

ion y(x). Assume that at position (x, y) = (0, 0) the tangent of the wire is horizontal and the bead passes this point with a given speed v0. What should y(x) be so that the horizontal speed remains v0 at all times (that is what is the shape of the wire)? One simple solution is y = 0, find the other solution. (Hint: It should
Physics
1 answer:
eduard3 years ago
7 0

Answer:

y = constant

Explanation:

Bodies moving on the surface of Earth are subject to gravity, and they have a potential and kinetic energy. If there is no friction, the sum of the kinetic and potential energy remains constant.

Since potential energy depends on height, changes in altitude affect potential energy. Going higher increases this energy, this is accompanied by a reduction of kinetic energy and speed (since kinetic energy is related to speed). If the body goes down potential energy is reduced, but kinetic energy and speed increase.

For speed to remain constant the kinetic energy must remain constant. For the kinetic energy to remanin constant, the potential energy must remain constant, and for the potential energy to remain constant the height must remain constant.

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Would sound travel faster in an oven or a freezer?
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6 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
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