Answer:
P = 450 J
Explanation:
Given that,
Mass of a child, m = 18 kg
The vertical distance from the top to the bottom of the slide is 2.5 metres.
The Gravitational field strength = 10 N/kg
We need to find the decrease in gravitational potential energy of the child sliding from the top to the bottom of the slide.
The formula for the gravitational potential energy is given by :
P = mgh
Substituting all the values,
P = 18 kg × 10 m/s² × 2.5 m
P = 450 J
Hence, the decrease in gravitational potential energy is 450 J.
Build up of pressure between tectonic plates .
What are your answer choices?
Answer:
a.) 1567.2 m/s
b.) 149.4 m/s
Explanation:
Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.
The x-component of the third part can be calculated by assuming that it moves in a positive x axis.
The third mass = 26 - ( 7.8 + 8.8)
The third mass = 26 - 16.6
The third mass = 9.4kg
since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion
26 x 350 = -8.8 x 640 + 9.4V
9100 = -5632 + 9.4V
9.4V = 9100 + 5632
9.4V = 14732
V = 14732/9.4
V = 1567.2 m/s
(b) y-component of the velocity of the third part will be
7.8 x 180 = 9.4 V
1404 = 9.4V
V = 1404/9.4
V = 149.4 m/s
Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s