Answer:
The family in the first period is the alkali metal family.
Line - A: (15/10) = 1.5 inch/second
Line - B: (0/10) = 0
Line - C: (10/10) = 1.0 inch/second
Line - D: (-25/20) = -1.25 inch/second
<em>Line-A</em> represents the greatest speed.
Answer:
a) W = - 318.26 J, b) W = 0
, c) W = 318.275 J
, d) W = 318.275 J
, e) W = 0
Explanation:
The work is defined by
W = F .ds = F ds cos θ
Bold indicate vectors
We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces
Let's use trigonometry to break down weight
sin θ = Wₓ / W
Wₓ = W sin 60
cos θ = Wy / W
Wy = W cos 60
X axis
How the body is going at constant speed
fr - Wₓ = 0
fr = mg sin 60
fr = 15 9.8 sin 60
fr = 127.31 N
Y Axis
N - Wy = 0
N = mg cos 60
N = 15 9.8 cos 60
N = 73.5 N
Let's calculate the different jobs
a) The work of the force of gravity is
W = mg L cos θ
Where the angles are between the weight and the displacement is
θ = 60 + 90 = 150
W = 15 9.8 2.50 cos 150
W = - 318.26 J
b) The work of the normal force
From Newton's equations
N = Wy = W cos 60
N = mg cos 60
W = N L cos 90
W = 0
c) The work of the friction force
W = fr L cos 0
W = 127.31 2.50
W = 318.275 J
d) as the body is going at constant speed the force of the tape is equal to the force of friction
W = F L cos 0
W = 127.31 2.50
W = 318.275 J
e) the net force
F ’= fr - Wx = 0
W = F ’L cos 0
W = 0
Explanation:
The object's weight on Earth is 5.0 N, so its mass is:
F = ma
5.0 N = m (10 m/s²)
m = 0.50 kg
The acceleration on the moon is g/6 = (10 m/s²) / 6 = 1.67 m/s².
The velocity it reaches after 3.0 seconds is:
v = at + v₀
v = (1.67 m/s²) (3.0 s) + (0 m/s)
v = 5 m/s
So the momentum is:
p = mv
p = (0.50 kg) (5 m/s)
p = 2.50 kg m/s
Answer:
The acceleration will also double
Explanation:
F = m*a
a = F/m
plugging in sample numbers to prove
a= 100/4 = 25
a = 200/4 = 50