Answer:
velocity = 62.89 m/s in 58 degree measured from the x-axis
Explanation:
Relevant information:
Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.
Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.
Before collision Momentum of A = 1000 x 100 = 
kg - m/s in the right direction.
Before collision Momentum of B = 2000 x 80 = 1.6 x kg - m/s in upward direction.
Mass of System of after collision = 1000 + 2000 = 3000 kg
Now applying the Momentum Conservation, we get
Initial momentum in right direction = final momentum in right direction =
And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x
So, 
= 
m/s
and 
m/s
Therefore, velocity is = 
= 
= 62.89 m/s
And direction is
tan θ = 
= 1.6
therefore, 
= 
from x-axis
Answer:
532 millimeters of mercury
Explanation:
In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:
1 atm = 760 mm Hg
Therefore, we can solve the problem by setting up the following proportion:
Solving for x, we find
Answer: Gamma rays
Explanation: The given waves belong to the electromagnetic spectrum which consists of different electromagnetic radiations arranged in terms of increasing wavelengths or decreasing frequencies.
Thus 
E= energy
= frequency
c = speed of light
= wavelength
Thus frequency and wavelength are inversely related. The waves having high energies ave high frequencies and have shorter wavelengths.
Thus gamma rays having highest energy have highest frequency and shortest wavelength.
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = 
/ W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
