Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer:
<em>Gravity</em><em>.</em><em> </em><em>The</em><em> </em><em>weight-force</em><em> </em><em>or</em><em> </em><em>weight</em><em> </em><em>of</em><em> </em><em>an</em><em> </em><em>object</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>because</em><em> </em><em>of</em><em> </em><em>Gravity</em><em>,</em><em> </em><em>which</em><em> </em><em>acts</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>object</em><em> </em><em>attracting</em><em> </em><em>it</em><em> </em><em>towards</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>earth</em><em>.</em>
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>
Answer:
1. the force which can be felt or act only when two objects are in contact is known as contact force.
for example: frictional force, muscular force,
tension, air resistance .
2. the force which can be felt or act even when two objects are in contact or not is known as non-contact force.
for example: magnetic force, gravitational force, electrostatic force.
Answer:
Explanation:
It is given that,
Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :
R = Rydberg constant,
Solving above equation we get the value of final n is,
or
So, it will relax in the n = 2. Hence, this is the required solution.