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
Answer:
d = 13 miles
Explanation:
Lets say the position of court house is origin in this case
her office is located at 4 miles west and 4 miles south of court house
so here we have coordinate of the office with respect to court house is given as

now the position of her home is located at 1 miles east and 8 miles north of the court house
so the coordinates of her home is given as

now the change in the position is given as the distance between office and home



The Kinetic energy would be 1/2IL².
<h3>What is
Rotational Kinetic energy ?</h3>
- Rotational energy also known as angular kinetic energy is defined as: The kinetic energy due to the rotation of an object and is part of its total kinetic energy. Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.
As we know linear Kinetic energy = 1/2mv²
where m= mass and v= velocity.
Similarly rotational kinetic energy is given by = 1/2IL²
where I- moment of inertia and L=angular momentum.
To know more about the Kinetic energy , visit:
brainly.com/question/29807121
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Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s

t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)