PHYSICS MIDDLE SCHOOL

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 the 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?

Answers

Answer 1

Answer:

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:

Work and Kinetic Energy

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

$\displaystyle k=(mv^2)/(2)$

We have

$\displaystyle \Delta k=(mv_1^2)/(2)-(mv_0^2)/(2)$

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

$\displaystyle W=(mv_1^2)/(2)-(mv_0^2)/(2)$

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

$\displaystyle k_o=(mv_0^2)/(2)$

$\displaystyle k_o=((75)6^2)/(2)$

$\boxed{\displaystyle k_o=1350\ J}$

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

$\displaystyle k_1=((75)0^2)/(2)$

$\boxed{\displaystyle k_1=0\ J}$

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

$\displaystyle F=(W)/(x)$

$\displaystyle F=(-1350\ J)/(2\ m)$

$\boxed{F=-675\ N}$

The negative sign indicates the force is against movement

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I traveled 6.00 km to school in 0.250 hours . What was my average speed

Answers

Answer: 6.67 m/s

Explanation:

Recall that average speed is the total distance covered by a moving body divided by the total time taken. The SI unit is metres per second.

i.e Average speed = distance/time taken

Given that:

Distance covered = 6.00 km

Convert kilometers to meters

If 1 km = 1000 m

6.00 km = (6.00 x 1000) = 6000m

Time taken to school = 0.250 hours

Convert time in hours to seconds

If 1 hour = 60 minutes & 1 minute = 60 seconds

Then, 0.250 hours = (0.250 x 60 x 60)

= 900 seconds

Average speed = ?

Average speed = 6000m/900s

= 6.67 m/s

Thus, your average speed to school is 6.67 m/s

Explain in your own words what the theory of evolution is.

Answers

The theory of evolution is that organisms evolve over time.

What object circles around a planet

Answers

A object that circles around a planet is called a satellite.

an object that circles around a planet is a satellite. earth's satellite is the moon.

Q: What happens when cold air approaches a body of warm air?

Optional Answers:

A. The Warm air rises.

B. The warm and cold air both disappear.

C. The warm and cold air mix immediately.

D. The cold air rises.

Answers

A. The warm air rises! Hope this helps

Answer a is correct.
Warm air rises.
EXAMPLE: when your cooking the hot air or warm air rises or moves upward.
Hope I helped.

10 POINTS!!! Determine the pressure of your book in pascals (Pa). Show your work! (the pressure of the book in psi is 0.03 psi) Also the weight of the book is 3 lbsHInt: 1 Pa = 1 N/m^2 Weight (force) = ma 1 kg = 2.2 lbs 1 inch = 2.54 cm 1m = 100 cm

Answers

You've listed a lot of data here, in both metric and customary units,
and I'm not even sure it's all needed. Let me try and boil it down:

Pressure on a surface =
   (total force on a surface) divided by (area of the surface).

The answer to the question is the pressure expressed in pascals.
There's actually enough information here to answer the question
in 2 different ways. We could ...

-- simply convert (0.03 pound per inch²) to pascals, or
-- go through the whole calculation of force, area, and then their quotient.

To me, converting 0.03 psi to Pa looks easier.

-- 1 pascal = 1 newton / 1 meter²

-- On Earth, 1 kilogram of mass weighs 9.8 Newtons and 2.2 pounds.
From this, we can calculate that

2.2 pounds of force = 9.8 newtons of force.

   1 pound = 4.45 newtons

(0.03 pound/inch²) x (4.45 newton/pound) x (1inch/2.54cm)² x (100cm/1m)² =

(0.03 x 4.45 x 1² x 100²) / (2.54² x 1²)    newton/meter² = 206.9 Pa .

A shopper pushes a 7.32 kg grocery cartwith a 14.7 N force directed at
-32.7° below the horizontal.
What is the normal force on the cart?

Answers

Answer:

The "normal force" on the "cart" 63.893 N.

Explanation:

To find normal force on the cart, use the equation

Normal force = mg + F sinx,

“m” being the object's mass,

“g” being the acceleration of gravity,

“x” being the angle of the cart

Given values

M = 7.33 kg

F = 14.7 N

$x=-32.7^(\circ)$

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^(2) \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^(2) \text { on Earth }$

Substitute the values in above equation

Normal force = (7.33 × 9.8) + 14.7 sin(-32.7°)

Normal force = 71.834 + 14.7 × (-0.5402)

Normal force = 71.834 - 7.94094

Normal force = 63.893 N

The "normal force" on "the cart" 63.893 N.

The normal force on the cart is 79.7 N

Explanation:

In order to find the normal force, we have to analyze the forces acting on the cart on the vertical direction.

In the vertical direction, we have the following forces:

The weight of the cart, downward, of magnitude , where m is the mass of the cart and g is the acceleration of gravity

The normal force on the cart, upward, we indicate it with N

The component of the pushing force acting in the vertical direction, downward, of magnitude , where F is the magnitude of the force and is the angle of the force with the horizontal

Therefore, the equation of the forces on the cart in the vertical direction is:

where the net force is zero since the cart is balanced in the vertical direction. We have:

We take the angle as positive since we are already considering the downward direction in the equation.

Substituting and solving for N, we find the normal force:

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

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