Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;
Therefore, his acceleration rate is -0.00186 m/s²
2.57 joule energy lose in the bounce
.
<u>Explanation</u>:
when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground
Formula for gravitational potential energy given by
Potential Energy = mgh
Where
,
m = mass
g = acceleration due to gravity
h = height
Potential energy when ball hits the ground
m= 0.375 kg
h = 1.37 m
g = 9.8 m/s²
Potential Energy = 5.03 joule
Potential energy when ball bounces up again
h= 0.67 m
Potential Energy = 2.46 joule
Energy loss = 5.03 - 2.46 = 2.57 joule
2.57 joule energy lose in the bounce
2 is the answer have a nice day <3
Answer:
See the explanation below.
Explanation:
We know that density is defined as the relationship between mass and volume.
where:
m = mass [kg]
V = volume [m³]
Therefore Ro is given in:
![[kg/m^{3} ]](https://tex.z-dn.net/?f=%5Bkg%2Fm%5E%7B3%7D%20%5D)
