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kkurt [141]
3 years ago
13

A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th

e runner before he begins his slide? b. What is the kinetic energy of the runner once he reaches the base? c. What is the change in the kinetic energy of the runner? d. How much work is done by friction in stopping the runner? e. If the runner slides for 2 m, what is the force of friction that acts upon him?
Physics
1 answer:
lana [24]3 years ago
6 0

Answer:

a. \displaystyle k_o=1350\ J

b. \displaystyle k_1=0\ J

c. \Delta k=-1350\ J

d. W=-1350\ J

e. F=-675\ N

Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

\Delta k=k_1-k_0

Knowing that

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The work done by the force who caused the change of velocity (acceleration) is

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

If we know the distance x traveled by the object, the work can also be calculated by

W=F.x

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

\displaystyle k_o=\frac{mv_0^2}{2}

\displaystyle k_o=\frac{(75)6^2}{2}

\boxed{\displaystyle k_o=1350\ J}

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

\displaystyle k_1=\frac{(75)0^2}{2}

\boxed{\displaystyle k_1=0\ J}

c. The change of kinetic energy is

\Delta k=k_1-k_0=0\ J-1350\ J

\boxed{\Delta k=-1350\ J}

d. The work done by friction to stop the player is

W=\Delta k=k_1-k_0

\boxed{W=-1350\ J}

e. We compute the force of friction by using

W=F.x

and solving for x

\displaystyle F=\frac{W}{x}

\displaystyle F=\frac{-1350\ J}{2\ m}

\boxed{F=-675\ N}

The negative sign indicates the force is against movement

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Answer:

72 km/h

Explanation:

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So, here to find the speed in km/h we will change the metre to km and seconds to hours .

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and,

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help kiya kya?

7 0
3 years ago
Read 2 more answers
Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

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F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
3 years ago
What type of weather would a continental Polar air mass bring
fgiga [73]

Answer:

Continental polar ( cp):

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6 0
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A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak
Sedbober [7]
By v = u - at 
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<span>=>a = 16km/hr/sec

I hope this helped!</span>
3 0
3 years ago
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<h3>On a roller coaster, what is centripetal force?</h3>

An item travelling in a circle is pushed inward toward what is known as the center of rotation, which is essentially what a roller coaster accomplishes when it travels through a loop. The force that maintains an object moving along a curved route is this pull toward the center, or centripetal force.

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