Answer:
0.006<357<700.003<6010<9256.0<9520.00
the answer is d they are essential to all ecosystems
Explanation:
Given parameters:
Distance = 15miles north = 24140.2m
Initial velocity = 0m/s
Final velocity = 4m/s
Unknown:
Speed, velocity and acceleration = ?
Solution:
The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.
Speed =
The speed of the student is 4m/s
Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;
Velocity =
The velocity of the student is 4m/s due north
Acceleration is the change in velocity with time;
To find the acceleration, we use
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
4² = 0² + 2x a x 24140.2
a =
= 0.00033m/s²
Answer:
5) 13 revolutions (approximately)
Explanation:
We apply the equations of circular motion uniformly accelerated :
ωf²= ω₀² + 2α*θ Formula (1)
Where:
θ : angle that the body has rotated in a given time interval (rad)
α : angular acceleration (rad/s²)
ω₀ : initial angular speed ( rad/s)
ωf : final angular speed ( rad/s)
Data:
ω₀ = 18 rad/s
ωf = 0
α = -2 rad/s² ; (-) indicates that the wheel is slowing
Revolutions calculation that turns the wheel until it stops
We apply the formula (1)
ωf²= ω₀² + 2α*θ
0 = (18)² + 2( -2)*θ
4*θ = (18)²
θ = (18)²/4 = 81 rad
1 revolution = 2π rad
θ = 81 rad * 1 revolution / 2πrad
θ = 13 revolutions approximately
Answer:
<em>The internal energy change is 330.01 J</em>
Explanation:
Given
the initial volume = 5.75 L
the final volume = 1.23 L
is the external pressure = 1.00 atm
q the heat energy removed = -128 J (since is removed from the system)
expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;
W = -
ΔV
W = -1.00 x(1.23 - 5.75)
W = -1.00 x -4.52
W = 4.52 L atm
converting to joules we have
W = 4.52 L atm x 101.33 J/ L atm = 458.01 J
The internal energy change during compression can be calculated thus;
ΔU = q + W
ΔU = -128 J + 458.01 J
ΔU = 330.01 J
Therefore the internal energy change is 330.01 J