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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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In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into t
Pie

Answer:

A) Q_a=74256\ J

B) Q=93562560\ J

Explanation:

Given:

  • temperature of air, T_a=-19+273=254\ K
  • temperature of lungs, T_l=37+273=310\ K
  • specific Heat exchanged from the lungs , c_l=0.47\ J.kg^{-1}.K^{-1}
  • specific heat of air, c_a=1020\ J.kg^{-1}.K^{-1}
  • mass of 1 L air, m'=1.3\ kg
  • breath rate, b=21\ breath.min^{-1}

A)

Now,

amount of heat needed to warm the air of lungs to the body temperature:

Q_a=m'.c_a.\Delta T

Q_a=1.3\times1020\times (310-254)

Q_a=74256\ J

B)

Amount of heat lost per hour:

<u>No. of breaths per hour:</u>

B=b.60

B=21\times 60

B=1260

<u>Now the total loss of energy in 1 hr.:</u>

Q=Q_a.B

Q=74256\times 1260

Q=93562560\ J

7 0
2 years ago
7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0
2 years ago
An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.
Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

\sin\theta=\dfrac{v'}{v}

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

\sin\theta=\dfrac{75}{310}

\theta=\sin^{-1}(\dfrac{75}{310})

\theta=14.0^{\circ}

We need to calculate the resultant speed

Using formula of resultant speed

\cos\theta=\dfrac{v''}{v}

Put the value into the formula

\cos14=\dfrac{v''}{310}

v''=\cos14\times310

v''=300.79\ km/h

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0
2 years ago
Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom
Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

B = B_o(-\hat k)

now the velocity is from bottom to top

so we have

v = v_o \hat j

now the force on the moving charge is given as

\vec F = q(\vec v \times \vec B)

now we have

\vec F = (-e)(v_o \hat j \times B_o(-\hat k))

\vec F = e v_o B \hat i

so force will be towards Right

7 0
2 years ago
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

6 0
2 years ago
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