Answer:

Explanation:
Given data
Time t=2.5 minutes=150 seconds
Distance A=1600 ft=487.68 m........east
Distance B=2500 ft=762m ........north
To find
Average velocity
Solution
First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse
So


Answer:
laws of motion relate an object’s motion to the forces acting on it. In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Answer:
The angular speed after 6s is
.
Explanation:
The equation

relates the moment of inertia
of a rigid body, and its angular acceleration
, with the force applied
at a distance
from the axis of rotation.
In our case, the force applied is
, at a distance
, to a ring with the moment of inertia of
; therefore, the angular acceleration is



Therefore, the angular speed
which is

after 6 seconds is


Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa