The first large herbivores and carnivores appeared during the Mesozoic Era, which is divided into three periods-- Triassic, Jurassic and Cretaceous. The herbivores and and carnivores increased in size during the Jurassic Period. And some of the largest dinosaurs emerged during this period as well.
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Answer:</h2>
(a) 6.95 x 10⁻⁸ C
(b) 6.25N/C
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Explanation:</h2>
The electric field (E) on a point charge, Q, is given by;
E = k x Q / r² ---------------(i)
Where;
k = constant = 8.99 x 10⁹ N m²/C²
r = distance of the charge from a reference point.
Given from the question;
E = 10000N/C
r = 0.250m
Substitute these values into equation(i) as follows;
10000 = 8.99 x 10⁹ x Q / (0.25)²
10000 = 8.99 x 10⁹ x Q / (0.0625)
10000 = 143.84 x 10⁹ x Q
Solve for Q;
Q = 10000/(143.84 x 10⁹)
Q = 0.00695 x 10⁻⁵C
Q = 6.95 x 10⁻⁸ C
The magnitude of the charge is 6.95 x 10⁻⁸ C
(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;
E = k x Q / r²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100
E = 6.25N/C
Therefore, at 10.0m, the electric field will be just 6.25N/C
This link should help you out!!
http://mszopiak.weebly.com/uploads/5/5/8/1/55815541/p_vtgraphskey_mszopiak.pdf
Answer:
a) Tc = 750 [N] ;b) See the explanation below.
Explanation:
To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.
b)
The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.
a)
In this way we can find by means of a sum of forces on the y axis equal to zero:
- 850 - 450 + 550 + Tc = 0
Tc = 750 [N]
Answer:
amount of energy = 4730.4 kWh/yr
amount of money = 520.34 per year
payback period = 0.188 year
Explanation:
given data
light fixtures = 6
lamp = 4
power = 60 W
average use = 3 h a day
price of electricity = $0.11/kWh
to find out
the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66
solution
we find energy saving by difference in time the light were
ΔE = no of fixture × number of lamp × power of each lamp × Δt
ΔE is amount of energy save and Δt is time difference
so
ΔE = 6 × 4 × 365 ( 12 - 9 )
ΔE = 4730.4 kWh/yr
and
money saving find out by energy saving and unit cost that i s
ΔM = ΔE × Munit
ΔM = 4730.4 × 0.11
ΔM = 520.34 per year
and
payback period is calculate as
payback period = 
payback period = 
payback period = 0.188 year