m = mass of the truck = 23 00 kg
v = speed of the truck down the highway = 32 m/s
K = kinetic energy of the truck = ?
kinetic energy of the truck down the highway is given as
K = (0.5) m v²
inserting the values
K = (0.5) (2300) (32)²
K = (0.5) (2300) (1024)
K = (1150) (1024)
K = 1177600 J
hence the kinetic energy of the truck comes out to be 1177600 J
Explanation:
We have,
The initial position of an object is zero.
The starting velocity is 3 m/s and the final velocity was 10 m/s.
The object moves with constant acceleration..
The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".
Answer:
3.49 seconds
3.75 seconds
-43200 ft/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Time the parachutist falls without friction is 3.19 seconds
Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity
So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds
Now the initial velocity of the last half height will be the final velocity of the first half height.
Since the height are equal
Time taken to fall the first half is 2.65 seconds
Total time taken to fall is 2.65+1.1 = 3.75 seconds.
When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.
Magnitude of acceleration is -43200 ft/s²
I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².
Taking downwards as positive, use v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s