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kondaur [170]
3 years ago
5

On a windless day, two identical balls P and Q are dropped from the top and the middle of a tower at exactly the same time as sh

own .Which will be more: the time taken by P to reach the middle of the tower ,or the time taken by Q to reach the ground?
Physics
1 answer:
Kruka [31]3 years ago
3 0

Answer: time is the same

Explanation: the distance(H) is the same in each case .

we drop the balls , no drag force using basic kimnematics

y =gt*t/2  , yo=0 , vo=0 , y=H  , so : t= sqrt(2H/g)

comment: if distance H starts to grow....we could begin to note a difference  because of gravity g is smaller as we go up

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At what angle two forces P + Q and (P - Q) act so that their resultant is :
stiv31 [10]

Use resultant formula

\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}

So

#1

A be p+q and B be p-q

\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}

\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2

\\ \rm\Rrightarrow 2cos\alpha=1

\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}

\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}

#2

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)

\\ \rm\Rrightarrow 2cos\beta=0

\\ \rm\Rrightarrow cos\beta=0

\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}

#3

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2

\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)

\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}

\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)

8 0
2 years ago
Read 2 more answers
Where would you expect to have more touch receptors: on the palm of your hand or on the back of your hand? Explain your reasonin
anygoal [31]

Answer:

ive answered this

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please check your previose question

6 0
3 years ago
What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the
Agata [3.3K]
The charge of the copper nucleus is 29 times the charge of one proton:
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the charge of the electron is
e=-1.6 \cdot 10^{-19}C
and their separation is
r=1.0 \cdot 10^{-12} m

The magnitude of the electrostatic force between them is given by:
F=k_e  \frac{Qe}{r^2}
where k_e is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
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3 years ago
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