It increases. Mercury takes 88 days to orbit the sun once. The Earth takes a year. Pluto takes 248 years.
Answer:
Amplitude—distance between the resting position and the maximum displacement of the wave
Frequency—number of waves passing by a specific point per second
Period—time it takes for one wave cycle to complete
wavelength λ - the distance between adjacent identical parts of a wave, parallel to the direction of propagation.
Tension - described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar one-dimensional continuous object, or by each end of a rod, truss member, or similar three-dimensional object
1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .
1 N = 1 kg-m/s² .
It's a force equal to roughly 3.6 ounces.
Answer: c. 1.3 m/s^2
Explanation:
When he is at rest, is weight can be calculated as:
W = g*m
where:
m = mass of the man
g = gravitational acceleration = 9.8m/s^2
We know that at rest his weight is W = 824N, then we have:
824N = m*9.8m/s^2
824N/(9.8m/s^2) = m = 84.1 kg
Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.
Then the new weight is calculated as:
W = m*(g + a)
and we know that in this case:
W = 932N
g = 9.8m/s^2
m = 84.1 kg
Then we can find the value of a if we solve:
932N = 84.1kg*(9.8m/s^2 + a)
932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a
11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2
The correct option is C
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
