The statement “When
an object is in orbit, it is falling at the same rate at which the Earth is
curving” is true. The speed of a satellite orbiting the earth depends only on
the mass of the earth and the mass of the satellite.
0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)
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Below are the choices:
only the x component
only the y component
both the x and y components
neither the x nor the y component
The answer is neither the x nor the y component
Answer:
P = (2 + 3) * V where V is their initial speed (total momentum)
P = 2 * 10 + 3 * Vx where Vx here would be V3
If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.
Also work depends on the sum of the velocities
W (initial) = 1/2 (2 + 3) V^2 the initial kinetic energy
W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2
It appears that more information is required for this problem
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]