In the equation below, the mass of the helium (He) atom is ________ the mass of the hydrogen (H) atoms combined.
1 answer:
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1) 12.5 N east
There are two forces acting on the box along the horizontal direction:
- The applied force of 15 N east, we indicate it with F
- The force of friction of 2.5 N west, we indicate it with
Taking east as positive direction, we can write the two forces has
Therefore, the net force on the box will be:
And the positive sign means the direction is east.
2)
We can solve this part by using Newton's second law:
where
is the net force on the box
m is its mass
a is the acceleration
For the box in this problem,
(east)
m = 5.0 kg
Solving for a, we find the acceleration:
And the direction is the same as the net force (east)
Answer:
At t = 15.0 s the magnitude of the velocity is 58.31 m/s
Explanation:
The given parameters are;
V₀ₓ = 30 m/s
= 100 m/s
The time of flight of the projectile = 25 s
For projectile motion;
Vₓ = V₀ × cos(θ₀)
The magnitude of the velocity V = √(V₀ₓ² + ²)
We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s
cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109
θ₀ = cos⁻¹(3/√109) = 73.3°
The components of the velocity after time t is given by the relations;
Vₓ = V₀ × cos(θ₀) = 30 m/s
= V₀ × sin(θ₀) - g×t
When = 0, we have;
0 = V₀ × sin(θ₀) - g×t
g×t = V₀ × sin(θ₀) = 10·√109×0.958 = 100 m/s
t = 100/g = 100/10 = 10 s
The time to reach maximum height = 10 s
At 15.0 seconds, we have;
= V₀ × sin(θ₀) - g×t = 10·√109×0.958 - 10×15 = -50 m/s
Therefore, the projectile is returning at 50 m/s
The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.