In navigation bearing refers to : 1. the direction of motion itself; 2. the direction of a distant object relative to the current course 3.the angle away from North of a distant point as observed at the current point.
There is a thing called absolute bearing, and it refers to the angle between the magnetic North/true North and an object. (e.g. an object of 0 degrees would be dead ahead, whereas and object of 180 degrees would be behind you)
Answer:
the heaviest barbell that could be lifted is 390.6kg
Explanation:
Hello!
To solve this question you must follow the following steps
1. Find the amount of energy that can transform the body into motion, this is achieved by multiplying the breakfast consumed by the percentage of energy conversion.

2. We use the equation that defines the work done by a body that has weight when it is lifted, this is defined by the product of mass by gravity by height.
W=mgh
where
W=work=6898.5J
m=mass
g=gravity=9.81m/s^2
h=height=1.8m
now we solve for mass, and use the values.

the heaviest barbell that could be lifted is 390.6kg
Answer:
2.4 m/s
Explanation:
Given:
Velocity of the object moving north = 2.1 m/s
Velocity of the river moving eastward = 1.2 m/s
The resultant velocity is the vector sum of the velocities of object and river.
Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.
The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.
The magnitude of the resultant velocity (R) is given as:

Therefore, the resultant velocity has a magnitude of 2.4 m/s.
Answer:
The frictional force acting on the block is 14.8 N.
Explanation:
Given that,
Weight of block = 37 N
Coefficients of static = 0.8
Kinetic friction = 0.4
Tension = 24 N
We need to calculate the maximum friction force
Using formula of friction force

Put the value into the formula


So, the tension must exceeds 29.6 N for the block to move
We need to calculate the frictional force acting on the block
Using formula of frictional force

Put the value in to the formula


Hence, The frictional force acting on the block is 14.8 N.
Let’s do this together!
Okay so the acceleration formula is vf-vi over time .
So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s
So the time is 5s so 9m/s divided into 5s is 1.8m/s/2
So the answer is 1.8m\s2