Travel in a straight path with constant velocity as the rock doesn't want to break its inertial frame.
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα= 
Senα= 
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
The distance covered by the body is 114.3 m
Explanation:
The work done by a force exerted on an object is given by

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the object in this problem, we have
F = 350 N is the force applied
is the work done
if we assume that the force is applied parallel to the motion of the object
Solving for d, we find the distance covered by the object:

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Answer:
The gravitational force on the elevator = 4500N
Explanation:
The given parameters are;
The force applied by the elevator, F = 4500 N
The acceleration of the elevator = Not accelerating
From Newton's third law of motion, the action of the cable force is equal to the reaction of the gravitational force on the elevator which is the weight, W and motion of the elevator as follows;
F = W + Mass of elevator × Acceleration of elevator
∴ F = W + Mass of elevator × 0 = W
F = 4500 N = W
The net force on the elevator is F - W = 0
The gravitational force on the elevator = W = 4500N.