Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;
- Mass of second block,
- Tension on first rope;
- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:
Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity 
)
For the First Rope
Total mass hanging on it; 
So Tension of the rope;
Therefore, the tension in the first rope is 147 Newton
For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;
Therefore, the tension in the second rope is 98 Newton
Learn More, brainly.com/question/18288215
False
If all other factors, such as medium, are kept the same, longitudinal waves tend to be stronger.
Answer:
-4*10⁴ units.
Explanation:
As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must still remain to be zero.
So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.
This is simple as power in watts is equal to joules per second so we can do 1500 joules divided by 30 seconds which equals 50 watts
