At rest, initial speed zero
x=v(initial) t+ 1/2 at^2
-1000m=0(10) + 1/2 a 10^2
-1000m=50a
a = -20 m/s^2
Answer: both hoops have the same kinetic energy at the bottom of the incline.
Explanation:
If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.
K1 + U1 = K2 + U2
If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then
K1 = 0 and U2 = 0
⇒ ΔK = ΔU = m g. h
If both inclines have the same height, and both hoops have the same mass m, the change in kinetic energy, must be the same for both hoops.
Answer:
B - Earth's path around the Sun
Explanation:
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
