The average snowfall of the four days was 6 inches.
What is snow?
When atmospheric water vapor freeze into small ice crystals and falling in small white flakes or lying on the ground as a white layer, it is called snow.
In the question it is said that,
Amount of snow received by the town on first day = 10 inches.
Amount of snow received by the town on second day = 3 inches.
Amount of snow received by the town on third day = 7 inches.
Amount of snow received by the town on fourth day = 4 inches.
Total amount of snow received by the town on these four days = ( 10 + 3+ 7 + 4) inches.
=24 inches.
Hence, average amount of snow received by the town on these four days = total amount of snow received ÷ 4
= ( 24 ÷ 4) inches.
= 6 inches.
Hence, the average snow received by the little town was 6 inches on those days.
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Answer:
The average is 6
Explanation:
To find the average, you have to first add all the numbers. After you add all the numbers up, then you divide by how many numbers there are. This one had 4 numbers so you divide by 4
As it was categorized in the physics category it must be the mechanical power.
Power is basically the amount of work done per unit time or in simpler words the amount of energy transferred per unit time.
The SI unit for power is Watt. Which is basically J/s or Nm/s as wor has the unit of Nm.
In electrical components power can also be categorized as the amount voltage supplied multiplied by the current passed through the component.
a. be a closer to the nucleus
b. be larger in size
c. hold more elections
d.have different shapes
e.have more nodes
Answer:
The orbitalswolud be larger in size and have more nodes than orbitals.
Explanation:
Given: orbitals would __________ than orbitals.
Since, the distance between nucleus and orbital increases when we electrons moves away fro m one orbital to another orbital.
So, The orbitals is far from the nucleus as compared with orbitals.
Therefore, The orbitals wolud be larger in size than orbitals.
Now, finding the number of nodes of an orbital is calculated as
Then, number of orbital in .
And number of orbital in .
Therefore, orbitals wolud have more nodes than orbitals.
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Answer :
(b) 4d orbitals would be larger in size than 3d orbitals
(e) 4d orbitals would have more nodes than 3d orbitals
Explanation :
As we move away from one orbital to another, the distance between nucleus and orbital increases. So, 4d orbitals would be far to the nucleus than 3d orbitals.
Hence, 4d orbitals would be larger in size than 3d orbitals.
Number of nodes is any orbital is n - 1 where, n is principal quantum number.
So, number of orbital in 4d is 3.
And number of orbital in 3d is 2.
So, options (b) and (e) are correct.
1.0 x 10^2 A
1.0 x 10^-10 A
4.0 x 10^0 A
Answer:
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:
Explanation:
Answer:
The pickup truck and hatchback will meet again at 440.896 m
Explanation:
Let us assume that both vehicles are at origin at the start means initial position is zero i.e. = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.
Truck:
= 33.2 m/s, a = 0 (since the velocity is constant), = 0
Using
s = 33.2t .......... eq (1)
Hatchback:
, = 0 m/s (since initial velocity is zero), = 0
Using
putting in the data we will get
now putting 's' value from eq (1)
which will give,
t = 13.28 s
so both vehicles will meet up gain after 13.28 sec.
putting t = 13.28 in eq (1) will give
s = 440.896 m
So, both vehicles will meet up again at 440.896 m.
Answer:
Explanation:
initial speed of the pickup truck (Up) = 33.2 m/s
acceleration of the pickup truck (ap) = 0
initial speed of the hatchback = 0
acceleration of the hatchback (ah) = 5 m/s^{2}
how far away (s) do the cars meet up again?
from the equations of motion distance covered (s) = ut +
distance covered by the pickup = ut +
where
distance covered by the hatchback = ut +
where
= ......equation 2
when the cars meet, they both would have covered the same distance, therefore
now that we have the time it takes for both cars to meet, we can put the value of the time into equation 1 or equation 2 to get the distance at which they meet
from equation 1
from equation 2