By building dams and levees . Hope this helps
In case of an object sitting at rest on another base, there are two equal and opposite forces – Normal force and the gravity.
Answer: Option A
<u>Explanation:
</u>
When an object is placed at rest position on another object, there is a force exerted by the surfaces of the two contact objects. This force is denoted as Normal Force.
When an object such as a box is placed on a shelf, its surface exerts a contact force on the base of the shelf- The Normal force directed upward. Meanwhile, the gravity stays at its action and tries to pull the box towards itself.
Both of these forces however are equal and opposite and therefore, there is zero net force on the box. That's why it remains at rest, holding on Newton's third law.
Answer:
nuclear energy is a energy that holds together of atoms
light energy is a kind of kinetic with the ability to make types of light visible to human eyes
Answer:
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
Explanation:
As the complete question is not given here ,the table of data is missing which is as attached herewith.
From the maximized equation of the utility function it is evident that
For the equity, here as
is percentage of the equity which is to be calculated
is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
is the risk aversion factor which is given as 4.
is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59
By substituting values.
So the weight of equity is 48.94%.
Now the weight of T bills is given as
So the weight of T-bills is 51.05%.
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest
