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storchak [24]
2 years ago
6

What is the mass of a baseball that has a kinetic energy of 105 J and is traveling at 10 m/s?​

Physics
1 answer:
kkurt [141]2 years ago
5 0

Answer:

\boxed {\boxed {\sf 2.1 \ kilograms}}

Explanation:

Kinetic energy is the energy an object possesses due to motion. The formula 1/2 the product of mass and the squared velocity.

E_k=\frac{1}{2} mv^2

We know the baseball's kinetic energy is 105 Joules. It is also traveling at a velocity of 10 meters per second. `

First, convert the units of Joules to make unit cancellation easier later in the problem. 1 Joule (J) is equal to 1 kilogram square meter per square second (kg*m²/s²). The baseball's kinetic energy of 105 J is equal to 105 kg*m²/s².

Now we know 2 values:

  • E_k= 105 \ kg*m^2/s^2
  • v= 10 \ m/s

Substitute these values into the formula.

105 \ kg*m^2/s^2= \frac{1}{2} m (10 \ m/s)^2

Now we need to solve for m, the mass. Solve the exponent.

  • (10 m/s)²= 10 m/s * 10 m/s = 100 m²/s²

105 \ kg *m^2/s^2 = \frac{1}{2} m (100 \ m^2/s^2)

Multiply on the right side.

105 \ kg *m^2/s^2 =  m (\frac{1}{2} * 100 \ m^2/s^2)

105 \ kg *m^2/s^2 =  m (50 \ m^2/s^2)

The variable, m, is being multiplied by 50 square meters per square second. The opposite of multiplication is division, so we divide both sides by that value.

\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}=  \frac{ m (50 \ m^2/s^2)}{50 \ m^2/s^2}

\frac {105 \ kg *m^2/s^2 }{50 \ m^2/s^2}= m

The units of square meter per square second will cancel out.

\frac {105 }{50} \ kg= m

2.1 \ kg=m

The mass of the baseball is <u>2.1 kilograms. </u>

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<span>Same procedure for 206Pb: </span>
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3 years ago
A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ =  34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

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Hence,  the work done by tension before the block goes up the incline = 292.97 J

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So average speed = 600/6 = 100 km/hr

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So the bus's average speed = 27.78 m/s or 100 km/hr.

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Explanation:

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