Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
Answer:
Mass = 386 kg
Explanation:
<u><em>Density = Mass / Volume</em></u>
Mass = Density × Volume
Where D = 19300 kg/m³ , V = 0.02 m³
<em>Putting the given in the above formula</em>
Mass = 19300 × 0.02
Mass = 386 kg
Answer:
![125\sqrt[4]{8}](https://tex.z-dn.net/?f=125%5Csqrt%5B4%5D%7B8%7D)
Explanation:
A number of the form

can be re-written in the radical form as follows:
![\sqrt[n]{a^m}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D)
In this problem, we have:
a = 1,250
m = 3
n = 4
So, if we apply the formula, we get
![1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}](https://tex.z-dn.net/?f=1%2C250%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B%281%2C250%29%5E3%7D)
Then, we can rewrite 1250 as

So we can rewrite the expression as
![=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B4%5D%7B%282%5Ccdot%205%5E4%29%5E3%7D%3D5%5E3%20%5Csqrt%5B4%5D%7B2%5E3%7D%3D125%5Csqrt%5B4%5D%7B8%7D)
If the three spoon touch nothing happens because they are all at room Temperature
Answer:
d = 61.75 m
Explanation:
Given that,
A ball droped from a building.
We need to find how fast is it traveling after falling 3.55 s.
As it is dropped, its initial velocity is equal to 0.
Let d is the distance it covers after falling 3.55 s.
We can use second equation of motion to find d.

Here, u = 0 and a =g

So, it will cover 61.75 m after falling 3.55 seconds.