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Yakvenalex [24]
3 years ago
12

When you catch a baseball, what kind of work do you do, negative or positive? By catching the baseball, what do you change? Does

the baseball gain or lose energy? What kind of energy?
Physics
2 answers:
Julli [10]3 years ago
6 0

Answer

1.negative work

2. kinetic energy and motion of ball

3. ball lost energy

4. kinetic energy

Explanation:

1:when you catch a ball that time their are two forces involved

a. force exerted by ball on your hand

b. Force exerted by you on ball  to stop it. Here the direction of force applied by you on the ball is in opposite direction to the motion of ball. it means direction of applied force and displacement of ball are in opposite direction it means work done by you on baseball is negative a.

2.When we catch a baseball that time we change its kinetic energy, force and its motion. A moving ball has kinetic energy and to stop the ball we have to change kinetic energy of ball. When we catch a baseball the kinetic energy of ball is transferred to our hand. When we catch a ball our hand moves in backward direction with ball because our hand gain energy.

3. Baseball lost energy and this energy is transferred to our hand.

4. Baseball has kinetic energy

bagirrra123 [75]3 years ago
3 0
Negative energy by catching it. Changes the force and movement of the baseball. Loses energy. Kinetic energy
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wlad13 [49]

Answer:

12m/s

Explanation:

v_f=v_o+at

Let's call the velocity that the car maintains for 10 seconds v_f_1, and the final velocity v_f_2.

v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s

Hope this helps!

5 0
3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
Mumz [18]

Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t

Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

6 0
3 years ago
Read 2 more answers
satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is close
GrogVix [38]

Answer:

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m

8 0
3 years ago
A pendulum bob is given velocity u =
Mama L [17]

Answer:

V = P = 0 m/s

Explanation:

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<u>V = P = 0 m/s</u>

7 0
3 years ago
A rocket, blasting into space, accelerates at 2 m/s^2 for 20 seconds. What is its speed at the end of 20 seconds?​
OLEGan [10]

Answer:

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Change in speed = (ending speed) - (beginning speed)

                           =     (3,600 mi/hr) - ( 0 )

                           =       3,600 mi/hr

                           =     same as  1 mile/second.

Acceleration  =  (1 mi/sec) / (10 sec)

                     =     0.1 mi/sec² .

But 1 mile = 5,280 ft,

so

                      0.1 mi/s²  =  528 ft/s²          

Explanation:

have a great day

4 0
3 years ago
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