Answer:
Acceleration a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Explanation:
For the truck to accelerate without losing its load.
Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.
Fa ≤ F(friction)
But;
Fa = mass × acceleration
Fa = ma
ma ≤ F(friction)
a ≤ (F(friction))/m ......1
Given;
Fa = mass × acceleration
Fa = ma
mass m = 800 kg
F(friction) = 2400 N
Substituting the given values into equation 1;
a ≤ F(friction)/m
a ≤ 2400N/800kg
a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Answer:
θ=142.9°
Explanation:
d=1 *r
angle ϕ= 37.1°
the line connecting pebble and target should be tangent to a circle so
cos(180-ϕ-θ)=
=
∴ θ=180-ϕ-
θ= 180-37.1-0
θ=142.9°
The question doesn't give us enough information to answer.
The answer depends on the mass of the object, how long the force
acts on the object, the OTHER forces on the object, and whether the
object is free to move.
-- If you increase the force with which you push on a brick wall,
the amount of work done remains unchanged, namely Zero.
-- If you push on a pingpong ball with a force of 1 ounce for 1 second,
the ball accelerates substantially, it moves a substantial distance, and
so the work done is substantial.
-- But if you push on a battleship, even with a much bigger force ...
let's say 1 pound ... and keep pushing for a month ... the ship accelerates
microscopically, moves a microscopic distance, and the work done by
your force is microscopic.
they are the only ones with a cell wall.
Answer:
Explanation:
Given a school bus.
Let say initially the school bus is traveling with speed "v"
Let assume mass of school bus is "m"
Then, the initial kinetic energy is
K.E_initial = ½mv²
Now, if the initial velocity is tripled,
Then, the new velocity is
v_new = 3v.
Note: the mass of the school does not change it is constant
Then, new kinetic energy is
K.E_new = ½m(v_new)²
v_new = 3v
Then,
K.E_new = ½m(3v)²
K.E_new = ½m × 9v²
K.E_new = 9 × ½mv²
Since K.E = ½mv²
Then,
K.E_new = 9 × K.E
So, the new kinetic energy will be 9 times the initial kinetic energy.
So, option D is correct
D. It will be nine times greater.
