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Lerok [7]
3 years ago
14

A 0.14-kg baseball is moving at 41 m/s. A 0.058-kg tennis ball is moving at 67 m/s. Which of the two balls has higher kinetic en

ergy?
Physics
1 answer:
lbvjy [14]3 years ago
6 0

Answer:

The kinetic energy of tennis ball is higher than that of baseball.

Explanation:

mass of baseball m_{b} = 0.14 kg

velocity of baseball v_{b} = 41 m/s

mass of tennis ball m_{t}  = 0.058 kg

velocity of tennis ball v_{t}= 67 m/s

To find: the kinetic energy of baseball KE_{b} and tennis ball KE_{t}

We know that the kinetic energy is given by the equation,

KE = \frac{mv^{2}}{2}

the kinetic energy of baseball KE_{b} = \frac{m_{b}v_{b}  ^{2}}{2}

                                                           = \frac{0.14 * 41^{2}}{2}

                                                           = 117.67 J

the kinetic energy of tennis ball KE_{t} = \frac{m_{t}v_{t}  ^{2}}{2}

                                                             = \frac{0.058 * 67^{2}}{2}

                                                             = 130.181 J

Hence KE_{t} > KE_{b}, the kinetic energy of tennis ball is higher than that of baseball

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BlackZzzverrR [31]

Answer:

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Other experiments determine that the material has a temperature of fusion of

fusion =235 °C and a temperature of vaporization of vapor=481 °C.

If the sample of material has a mass of =8.60 g, calculate the specific heat when this material is a solid, and when it is liquid, l

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2 years ago
Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a
DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0
3 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

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change in enthalapy  5.07*10^5 J kg^{-1}

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gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

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change in internal energy = Cv(T_2 -T_1)

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change in enthalapy \Delta H = Cp(T_2 -T_1)

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change in entropy

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\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

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