Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
<h3>a. The impulse</h3>
The impulse is 100.0 Ns
The impulse I = Ft where
- F =average force = 50.0 N and
- t = time = 2.0 s
Substituting these values into the equation, we have
I = Ft
I = 50.0 N × 2.0 s
I = 100.0 Ns
The impulse is 100.0 Ns
<h3>b. Change in momentum</h3>
The change in momentum is 100 kgm/s
Since change in momentum Δp = I where I = impulse.
Since I = 100.0 Ns,
Substituting this into the equation, we have
Δp = I
= 100.0 Ns
= 100 kgm/s
The change in momentum is 100 kgm/s
<h3>c. Mass's change in velocity</h3>
The change in velocity is 25.0 m/s
Since change in momentum Δp = mΔv where
- m = mass = 4.0 kg and
- Δv = change in velocity.
Making Δv subject of the formula, we have
Δv = Δp/m
Substituting the values of the variables into the equation, we have
Δv = Δp/m
Δv = 100.0 kgm/s/4.0 kg
Δv = 25.0 m/s
The change in velocity is 25.0 m/s
Learn more about impulse here:
brainly.com/question/25700778
At maximum height, the velocity of the ball is 0.
v = u + at
0 = 34 - 32.2t
t = 1.06 seconds
In H(t), s = 0 because the ball is initially on the ground.
H(t) = -16(1.06)² + 34(1.06)
H(t) = 18.1 feet
The problem should only have one part to it, but this one has two.
Before I can do the mass/energy conversion, I have to go and
look up the proton mass for myself ... go out and collect the straw
to make my bricks, as it were. As if the fabulous bounty of 7 points
makes it worth it. They make us do everything around here.
OK. In my Physics book⁽¹⁾, the proton rest mass is
1.67 x 10⁻²⁷ kg.
The formula that relates mass to the equivalent energy is
E = m c² .
The method of applying the formula is known as "plug in what you know",
as follows:
E = (mass) x (speed of light)²
= (1.67 x 10⁻²⁷ kg) x (3 x 10⁸ m/s)²
= (1.67 X 10⁻²⁷ Kg) x (9 x 10¹⁶ m²/s²)
= (1.5 x 10⁻¹⁰) (kg-m²/s²)
= 1.5 x 10⁻¹⁰ joule .
____________________________________
⁽¹⁾ Halliday, David and Resnick, Robert, Physics , John Wiley & Sons,
Inc., 1960, inside front cover, "SELECTED PHYSICAL CONSTANTS".
Answer:
the molrcular mass of carbon dioxide id 44.01amu
Explanation:
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