<span>Forward & falling. Hope this helps!</span>
<u>We are given:</u>
The force applied on the poor hamster (F) = 12 N
Acceleration of the poor Hamster (a) = 8 m/s²
<u>Solving for the mass of the Poor Hamster:</u>
From newton's second equation of motion, we know that:
F = ma
<em>replacing the given values</em>
12 = 8 * m
m = 12/8 kg
m = 3/2 kg
The poor Hamster weighs 3/2 kg
Answer:
P= 0.87g/mL or 0.87g/cm^3
Explanation:
P=m/v
P=density
P=17.4g/20mL
P= 0.87g/mL
1mL=1cm^3
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
Answer is: <span>the objects potential energy is 24500 J.
</span>m(object) = 50 kg.
h(object) = 50 m.
g = 9,8 m/s².
E(object) = m·g·h.
E(object) = 50 kg · 9,8 m/s² · 50 m.
E(object) = 24500 N·m = 24500 J = 24,5 kJ.
g - <span>the acceleration of free fall.
mg - </span><span>weight of the object.</span>
