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GREYUIT [131]
2 years ago
9

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

Physics
1 answer:
sveta [45]2 years ago
6 0

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with m_{Pa} and its initial velocity with u_{Pa}

Let represent the mass of Ricardo with m_{Ri} and its initial velocity with u_{Ri}

At rest ;

their velocities will be zero, i.e

u_{Pa} = u_{Ri} = 0

The initial momentum for this process can be represented as :

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} = 0

after push off from each other then their final velocity will be v_{Pa} and v_{Ri}

The we can say their final momentum is:

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri} = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} =  m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

Since the initial velocities are stating at rest then ; u = 0

m_{Pa}(0) + m_{Pa}(0) = m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}  = 0

m_{Pa}v_{Pa} = - m_{Ri}v_{Ri}

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So m_{Ri} > m_{Pa} ;

Then \mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

m_{Pa}v_{Pa} =  m_{Ri}v_{Ri}

The ratio is

\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1

v_{Pa} >v_{Ri}

Therefore, Paula will have a greater speed than Ricardo after the push-off.

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Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
1 year ago
A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidentally
vazorg [7]

Answer:

a. v1 = 5.06 m/s,  v2 = 3.96 m/s ,  R = 1.27

b. t = 1 hr, 11 min, 26 sec  

Explanation:

Using the Bernoulli's laws to use the conserved energy

a. Solve the speed and the radio of this speed of the tank is open to the air

p₀ + ρgh₀ + ½ρv₀² = p₁ + ρgh₁ + ½ρv₁²

5000Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 25.68  m²/s²

v1 = 5.06 m/s

Because it is open the tank so P=0 pa so:

0 Pa + 1000kg/m³ * 9.8m/s² * 0.800m + 0 = 0 + 0 + ½ * 1000kg/m³ * v²

v² = 15.68  m²/s²

v2 = 3.96 m/s

The ratio on the air is solve using both velocities so:

R = v1/v2 = 5.06 m/s / 3.96 m/s

R = 1.27

b. Now to find the time it takes for the tank to drain if the tank is open to the air

dh/dt = -u

dh/dt = -v * A/A'

dh/dt = v*(.02m)²/(2.0m)² = -v / 10000

and we can further substitute for v:

dh/dt = -(1/1e4)*√[(p+9800h)/500]

Solve replacing

-(1000/49)*√(49000h) = t + C

-(1000/49)*√(49000*0.8) = 0 + C

C = - 4040.6

Then when h = 0,

t = 4286 s

t = 1 hr, 11 min, 26 sec  

6 0
2 years ago
A marble is dropped straight down from a distance h above the floor.
Mrac [35]
Fm=Fe and am>ae
Hopefully this helps
7 0
2 years ago
Whennes
rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

v=v_{0}+g*t

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

v=v_{o}+g*t

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

y =   0.5*g*t^{2}

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]

4 0
2 years ago
A solid cylinder of mass 12.0 kg and radius 0.250 m is free to rotate without friction around its central axis. If you do 75.0 J
faltersainse [42]

Answer:

20 rad/s

Explanation:

mass, m = 12 kg

radius, r = 0.250 m

Moment of inertia of cylinder, I = 1/2 mr²

I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2

Work done = Change in kinetic energy

Initial K = 0

Final K = 1/2 Iω²

W = 1/2 Iω²

ω² = 2W/ I = 2 x 75 / (0.375)

ω = 20 rad/s

Thus, the final angular velocity is 20 rad/s .

8 0
1 year ago
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