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olya-2409 [2.1K]
3 years ago
12

a body of mass 3kg is dropped from a height 15m.what would be the velocity of the body at 10m of height? take g=10m/s​

Physics
1 answer:
mixer [17]3 years ago
5 0

Answer:

Explanation:

using v^2 = u^2 +2AS                                                                                                             here : v = ? , u = 0 , a = 10 , s = 10.                                                                                                            v*v = 0 + 2 (100)                                                                                                                    v^2 = 200                                                                                                                         v =  \sqrt{200                                                                                                            therefore , v = 14.14 m/s

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If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?
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The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

R=A=B

So, using the law of cosines, we have:

R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°

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An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh
lara [203]

Answer:

a_y = 4.9\ m/s^2

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}

\dfrac{dz}{dy}=-0.5

The differential form with the acceleration

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0.5=\dfrac{a_y}{0 + 9.8}

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a_y = 4.9\ m/s^2

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