Answer:
2.667m/s to the north and 3.333 m/s to the west
Explanation:
According to law of momentum conservation, the total momentum should be conserved before and after the explosion.
Before the explosion, the momentum was
0.5*2 = 1 kg m/s to the west
Therefore the total momentum after the explosion should be the same horizontally and vertically.
Vertically speaking, it was 0 before the explosion. After the explosion:
0.2*4 + 0.3v = 0
0.3v = -0.8
v = -0.8/0.3 = -2.667 m/s
So the vertical component of the 0.3kg piece is 2.667m/s to the north
Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:
0.2*0 + 0.3V = 1
0.3V = 1
V = 1/0.3 = 3.333 m/s
So the horizontal component of the 0.3kg piece is 3.333 m/s to the west
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
<h2>
Answer:</h2>
(a) 3.96 x 10⁵C
(b) 4.752 x 10⁶ J
<h2>
Explanation:</h2>
(a) The given charge (Q) is 110 A·h (ampere hour)
Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;
=> Q = 110A·h
=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]
=> Q = 110 x A x 3600s
=> Q = 396000A·s
=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C
Therefore, the number of coulombs of charge is 3.96 x 10⁵C
(b) The energy (E) involved in the process is given by;
E = Q x V -----------------(i)
Where;
Q = magnitude of the charge = 3.96 x 10⁵C
V = electric potential = 12V
Substitute these values into equation (i) as follows;
E = 3.96 x 10⁵ x 12
E = 47.52 x 10⁵ J
E = 4.752 x 10⁶ J
Therefore, the amount of energy involved is 4.752 x 10⁶ J
Answer:
the object will travel 0.66 meters before to stop.
Explanation:
Using the energy conservation theorem:
The work done by the friction force is given by:
![W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%5C%5CW_f%3D%5C%C2%B5%2Am%2Ag%2Ad%5C%5CW_f%3D0.35%2A4%2A9.81%2Ad%5C%5CW_f%3D13.7d%5BJ%5D)
so:
