Answer:
a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
Explanation:
a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:
Jet
(1)
Glider
(2)
Where:
- Thrust of jet engines, measured in newtons.
- Tension in the cable connecting the jet and glider, measured in newtons.
, - Masses of the glider and the jet, measured in kilograms.
- Acceleration of the glider-jet system, measured in meters per square second.
If we know that , and , then the solution of this system of equations:
By (2):
By (1):
The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
Answer:
A) 30 N upward
Explanation:
There are two forces on the block: buoyant force up and weight down. The net force is 5 N up, so:
5 N = B − W
The weight is 25 N so:
5 N = B − 25 N
B = 30 N
The buoyant force is 30 N up.
<h2>
Answer:Third</h2>
Explanation:
Newton's third law states that every action force has a equal and opposite reaction force.
Example 1:
Consider yourself hitting a wall with your hand,you are exerting a force on the wall but still you feel hurt because the wall exerts same force on your hand in opposite reaction.
Example 2:
Consider yourself standing on ground.You are exerting force on ground.But how are you able to stand on ground even if your mass is forcing you to go down?The answer is normal reaction force.The ground exerts a force on your feet which makes you to stand still.
Answer:
Two possible points
<em>x= 0.67 cm to the right of q1</em>
<em>x= 2 cm to the left of q1</em>
Explanation:
<u>Electrostatic Forces</u>
If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude
We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.
Equating
Operating and simplifying
To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.
Assuming the positive sign
:
Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative
The negative sign of x means q3 is located to the left of q1 (assumed in the origin).
Ammonia is a weak base because its nitrogen atom has an electron pair that readily accepts a proton. Also, when dissolved in water, ammonia acquires hydrogen ions from water to produce hydroxide and ammonium ions. It is the production of these hydroxide ions that imparts ammonia its characteristic basicity.