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3 years ago

Explain an amusement park you've ridden discuss the conversion of energy on that ride

1 answer:

6 0

Roller coaster...

going up... building the potential energy....

SCREAM... AND SCREAM... kinetic energy upon kinetic energy being used

rinse and repeat

You might be interested in

What is an unbalanced force

Butoxors [25]

There's no such thing as "an unbalanced force".

If all of the forces acting on an object all add up to zero, then we say that
the group of forces is balanced. When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all.

An example:
Two people with exactly equal strength are having a tug-of-war. They pull
with equal force in opposite directions. Each person is sweating and straining,
grunting and groaning, and exerting tremendous force. But their forces add up
to zero, and the rope goes nowhere. The group of forces on the rope is balanced.

On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal. The group of forces is unbalanced, and the rope moves.

A group of forces is either balanced or unbalanced. A single force isn't.

7 0

3 years ago

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

3 0

3 years ago

Consider a situation where you are playing air hockey with a friend. The table shoots small streams of air upward to keep the pu

artcher [175]

When we hit the puck from tap the puck will move forward.

This is due to the impulse provided by us at the time of hit. Due to this impulse the puck will move forward and start moving in some direction.

As soon as puck move forward the force on it is zero as the weight of the puck is counterbalanced by the air stream force and there is no other force on it so puck will continue its motion till it will hit at some other point.

So here the motion of the puck will be uniform motion till it will collide with some other points.

So here the correct option will be given as

moves with a constant speed until hitting the other end.

5 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$