Answer:
point of support on which a lever rotates.
Explanation:
The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.
The fulcrum provides the platform for a lever to torque.
- The force that opposes motion by the applied force is termed the frictional force.
- Friction is a force that opposes motion.
- The stored energy of an object is its potential energy.
- The potential energy is the energy due to the position of a body.
- The distance an object moves when doing work is termed its displacement.
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.
At its highest point, its velocity changed from upward to downward.
At that instant, its velocity was zero.
The acceleration of gravity is 9.8 m/s². That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second.
-- If the object is falling downward, it moves 9.8 m/s faster every second.
-- If the object is tossed upward, it moves 9.8 m/s slower every second.
The ball took 4 seconds to lose all of its upward speed. So it must have
been thrown upward at (4 x 9.8 m/s) = 39.2 m/s .
(That's about 87.7 mph straight up. Somebody had an amazing pitching arm.)
Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So 
Momentum = (mv).
<span>(2110 x 24) = 50,640kg/m/sec. truck momentum. </span>
<span>Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec</span>
Answer:
Explanation:
Velocity of plane relative to ground V_pg = ?
Given the velocity in vector form ,
velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j
V_wg = 60 i
V_pg = V_pw +V_wg
= 120 cos30 i + 120sin30j + 60i
= 164 i + 60 j
magnitude
=251 km / h
=