Answer:
The answer to your question is given below.
Explanation:
Mechanical advantage (MA) = Load (L)/Effort (E)
MA = L/E
Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)
VR = l/e
Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100
Efficiency = Wd/Wp x100
Recall:
Work = Force x distance
Therefore,
Work done by machine (wd) = load (L) x distance (l)
Wd = L x l
Work put into the machine (Wp) = effort (E) x distance (e)
Wp = E x e
Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)
Efficiency = Wd/Wp x100
Efficiency = (L x l) / (E x e) x 100
Rearrange
Efficiency = L/E ÷ l/e x 100
But:
MA = L/E
VR = l/e
Therefore,
Efficiency = L/E ÷ l/e x 100
Efficiency = MA ÷ VR x 100
Efficiency = MA / VR x 100
Answer:
2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.
Explanation:
The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:
![2Mg(s)+O_2(g)\rightarrow 2MgO(s)](https://tex.z-dn.net/?f=2Mg%28s%29%2BO_2%28g%29%5Crightarrow%202MgO%28s%29)
Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.
![[Mg]=1s^22s^22p^63s^2](https://tex.z-dn.net/?f=%5BMg%5D%3D1s%5E22s%5E22p%5E63s%5E2)
In order to attain noble gas configuration it will loose two electrons.
![[Mg]^{2+}=1s^22s^22p^6](https://tex.z-dn.net/?f=%5BMg%5D%5E%7B2%2B%7D%3D1s%5E22s%5E22p%5E6)
...[1]
Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.
![[O]=1s^22s^22p^4](https://tex.z-dn.net/?f=%5BO%5D%3D1s%5E22s%5E22p%5E4)
In order to attain noble gas configuration it will gain two electrons.
![[O]^{2-}=1s^22s^22p^6](https://tex.z-dn.net/?f=%5BO%5D%5E%7B2-%7D%3D1s%5E22s%5E22p%5E6)
..[2]
2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.
They are falling under the sole influence of gravity all objects<span> will </span>fall<span> with the </span>same<span> rate of </span><span>acceleration needless of there size</span>
Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,
![f=\frac{c}{\lambda}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bc%7D%7B%5Clambda%7D)
substituting the values in the equation we get,
![f=\frac{3.0\times 10^8 m/s}{2.9m}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B3.0%5Ctimes%2010%5E8%20m%2Fs%7D%7B2.9m%7D)
f = 1.03 x 10⁸Hz
Now,
The time period (T) = ![\frac{1}{f}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D)
or
T =
= 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target =
= 145 m