Answer:
Explanation:
mass of elephant, m1 = 5240 kg
mass of ball, m2 = 0.150 kg
initial velocity of elephant, u1 = - 4.55 m/s
initial velocity of ball, u2 = 7.81 m/s
Let the final velocity of ball is v2.
Use the formula of collision
v2 = - 16.9 m/s
The negative sign shows that the ball bounces back towards you.
(b) It is clear that the velocity of ball increases and hence the kinetic energy of the ball increases. This gain in energy is due to the energy from elephant.
Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N - =
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
Answer:
f.The period is independent of the suspended mass.
Explanation:
The period of a pendulum is given by
where
L is the length of the pendulum
g is the acceleration due to gravity
From the formula, we see that:
1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length
2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation
So, the only correct statements are
f.The period is independent of the suspended mass.
Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.
Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W