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Lerok [7]
3 years ago
15

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 43.5 m / s . Suppose

the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. What would the final velocities of the goalie and the puck be in this case
Physics
1 answer:
AnnZ [28]3 years ago
7 0

Answer:

0.06824 m.s for each  - puck and goalie

Explanation:

Mass of puck=m1= 0.11 kg

Mass of goalie = m2 = 70 kg

Initial vel of puck=V1i= 43.5 m/s

Initial vel of goalie = Vi2= 0 m/s

V1f = final velocity of puck

Vf2= final velocity of goalie

Because goalie hits the puck , the final velocity of both is equale i-e Vf1=Vf2=V

Using constant momentum rule;

m1V1i+m2V2i = m1 V1f+m2V2f

==> m1 V1i +0 = m1 V1f+m2V2f

==>m1 V1i  = (m1+m2) V

==> V= m1V1i ÷ (m1+m2) = 0.11 × 43.5 ÷ (0.11+70)= 4.785÷ 70.11 m/s = 0.06824 m/s

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Which of the following objects exerts the greatest gravitational force on the Earth?
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Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
A 1.80-m string of weight 0.0126 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
Veseljchak [2.6K]

Answer:

W = 0.135 N

Explanation:

Given:

- y (x, t) = 8.50*cos(172*x -2730*t)

- Weight of string m*g = 0.0126 N

- Attached weight = W

Find:

The attached weight W given that Tension and W are equal.

Solution:

The general form of standing mechanical waves is given by:

                            y (x, t) = A*cos(k*x -w*t)  

Where k = stiffness and w = angular frequency

Hence,

                           k = 172 and w = 2730

- Calculate wave speed V:

                            V = w / k = 2730 / 172 = 13.78 m/s

- Tension in the string T:

                            T = Y*V^2

where Y: is the mass per unit length of the string.

- The tension T and weight attached W are equal:

                           T = W = Y*V^2 = (w/L*g)*V^2

                            W = (0.0126 / 1.8*9.81)*(13.78)^2

                            W = 0.135 N

4 0
3 years ago
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