- Magnitude: 12.1 N.
- Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>
Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.
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The sum of forces on each direction will be the resultant force on that direction:
- Resultant force parallel to the 8 N force:
. - Resultant force normal to the 8 N force:
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Apply the Pythagorean Theorem to find the magnitude of the resultant force.
(3 sig. fig.).
The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:
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Find the size of the angle using inverse tangent:
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In other words, the resultant force is 17.0° relative to the 8 N force.
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
The answer is Convoluted endoplasmic reticulum
Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;

where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.