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notsponge [240]
2 years ago
12

A fully dressed person is at rest in the middle of a pond on perfectly frictionless ice and must get to shore. How can this be a

ccomplished? Assuming you've answered them in terms of momentum conservation, answer them also in terms of Newton's third law (or vice versa, if you answered already in terms of Newton's third law).
Physics
1 answer:
Paul [167]2 years ago
3 0

I'll explain it in both conservation of momentum and newtons third law.

Conservation of momentum : The total momentum of a closed system is always a constant

Newtons third law : An action has an equal and opposite reaction

Now if you're fully dressed on a frictionless pond at rest and you want to go to shore you have to make a sacrifice. Take off your clothes and throw them in the direction you DONT want to go. So the opposite side of the shore. In the case of conservation of momentum, the intial momentum is zero thus the total of the final momentum should be zero too. Pi = Pf. Once you throw your clothes away your clothes' momentum will be (Mc)(-Vc) and your momentum will be (M)(V) adding those two HAS to result in a zero because your initial momentum was zero.

Now explaining it in terms of Newtons third law, if you throw your clothes away with a force, the clothes will exert the same exact force on you. Since the ice is frictionless that force will cause you to accelerate. Now the acceleration depends on your mass and the force you exerted on the clothes.

Hope it makes sense!

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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
2 years ago
Suppose that, instead of the Coulomb force law, one finds experimentally that the force between any two charge q1 and q2 is Writ
denpristay [2]

Answer: E= KQ/r^2

Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.

The magnitude of an electric field E, at a point is given by Coulombs law as

E/ F/q

Where F= Coulombs force exertedon the charge and q= electric charge

E= F/q=(KQq)/r^2q

E=KQ/r^2

6 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den
Ymorist [56]

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

4 0
2 years ago
Read 2 more answers
Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD
NeX [460]

Answer:

KE= 1/2mv²

Explanation:

The kinetic energy of a body is the energy possessed by virtue of the body in motion

Given the parameters

m which is the mass of the body

v which is the velocity of the body too

K.E = kinetic energy

The expression for the kinetic energy of a body is given as

KE= 1/2mv²

3 0
2 years ago
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