Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s
charged objects will either attract or repel other charged objects
Explanation:
It is given that,
Mass of the object, m = 1.8 kg
Horizontal force acting on the object, F = 17 N
The object is pulled 0.2 meters from its equilibrium position. It is the amplitude of oscillation.
(a) We know from the Hooke's law that the force in simple harmonic motion is given by :
k = 85 N/m
(b) Let is the frequency of the oscillations. It is given by :
(c) Let v is the maximum speed of the object. In SHM, the maximum speed of the object is given by :
(d) At mean position, the speed of object is maximum. It means at x = 0, the speed of the object is maximum.
Hence, this is the required solution.
Answer:
The density of the block is 7.4g/ml.
Explanation:
We can determine the volume of the metal block by taking the difference between the volumes measured in the graduated cylinder:
Now, as we know that the average density of an object is calculated dividing its mass by its volume, we can calculate the density ρ of the metal block using the expression:
Finally, it means that the density of the metal block is 7.4g/ml.
Answer: A: The Mass of Object AND D. The amount of motion it’s particles have