Answer:
The magnitude of the impulse is 1.33 kg m/s
Explanation:
please look at the solution in the attached Word file
Answer:5.7m/s
Explanation:
Mass=1kg
Initial velocity=u=8m/s
height=h=1.6m
Final velocity =v
Acceleration due to gravity=g=9.8m/s^2
v^2=u^2-2xgxh
v^2=8^2-2x9.8x1.6
v^2=8x8-2x9.8x1.6
v^2=64-31.36
v^2=32.64
Take the square root of both sides
√(v^2)=√(32.64)
v=5.7
Speed at the height of 1.6m is 5.7m/s
Is the sciencetjfic law and security theory the same?
TRUE
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
All elements are balanced. There are 1 Mg, 1 O, 2 Li's and 2 Cl's.