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2 years ago

A 8kg cat is running 4m/s. how much kinetic energy does it have?

2 answers:

7 0

The formula for kinetic energy is, ${ E }_{ k }=\frac { 1 }{ 2 } \cdot m\cdot { v }^{ 2 }$
where m is mass and v is velocity.

Let's plug the given values in the formula.

${ E }_{ k }=\frac { 1 }{ 2 } \cdot m\cdot { v }^{ 2 }\\ \\ { E }_{ k }=\frac { 1 }{ 2 } \cdot 8\cdot { 4 }^{ 2 }\\ \\ { E }_{ k }=\frac { 1 }{ 2 } \cdot 8\cdot 16\\ \\ { E }_{ k }=4\cdot 16\\ \\ { E }_{ k }=64\quad \\ \\$

It's kinetic energy is 64 Joules.

olya-2409 [2.1K]2 years ago

6 0

Kinetic energy = (1/2) (mass) (speed)²

Kinetic energy = (1/2) (8 kg) (4 m/s)² =

      (4 kg) (16 m²/s²) =

   64 (kg-m/s²) (m) = 64 newton-m = <em>64 joules</em>

And may I say . . . that is one fat cat !   (about 17.6 pounds)

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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w

ioda

Answer:

$Q_{week} = 458884.6\, BTU$

Explanation:

The weekly water consumption of Kimonoski is:

$m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{bath.week} = 948.205\,lbm$

$m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)$

$m_{others, week} = 1346.218\,lbm$

$m_{week} = m_{bath,week} + m_{others, week}$

$m_{week} = 2294.423\,lbm$

The total energy required per week for hot water is:

$Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T$

$Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)$

$Q_{week} = 458884.6\, BTU$

3 0

1 year ago

A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of th

jekas [21]

Answer:

clockwise

Explanation:

According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.

Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.

The clockwise induced current will produce magnetic field in to the screen.

6 0

1 year ago

A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

6 0

1 year ago

A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet

kogti [31]

The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.

4 0

2 years ago

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