Answer:
d = <23, 33, 0> m
, F_W = <0, -9.8, 0>
, W = -323.4 J
Explanation:
We can solve this exercise using projectile launch ratios, for the x-axis the displacement is
x = vox t
Y Axis
y =
t - ½ g t²
It's displacement is
d = x i ^ + y j ^ + z k ^
Substituting
d = (23 i ^ + 33 j ^ + 0) m
Using your notation
d = <23, 33, 0> m
The force of gravity is the weight of the body
W = m g
W = 1 9.8 = 9.8 N
In vector notation, in general the upward direction is positive
W = (0 i ^ - 9.8 j ^ + 0K ^) N
W = <0, -9.8, 0>
Work is defined
W = F. dy
W = F dy cos θ
In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º
Cos 180 = -1
W = -F y
W = - 9.8 (33-0)
W = -323.4 J
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
Technically, this delivers a lot of energy into the Earth, but it’s
spread out over a large enough area that it doesn’t do much more than
leave footprints in a lot of gardens. A slight pulse of pressure spreads
through the North American continental crust and dissipates with little
effect. The sound of all those feet hitting the ground creates a loud,
drawn-out roar which lasts many seconds.
<span>The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.
So there will be only 2 electrons in the outer most shell.
According to the information mentioned above we can conclude the </span><span>unknown element likely belongs to the second group.
</span><span>I2 = 1752 kj/mol</span>