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a_sh-v [17]
3 years ago
12

A rock at rest falls off a tall cliff and hits the valley below after 3.5s. What is the rocks velocity as it hits the ground

Physics
2 answers:
pogonyaev3 years ago
7 0
T = 3.5 secs

Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
Paraphin [41]3 years ago
5 0
V = u + at
u=0(rest) ; a=gravitational acc(g)=10
v = 0 + 10(3.5) = 35m/s


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Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul
Gnom [1K]

Answer:

t=2.5\times 10^{-14}\ s

Explanation:

We know that charge on electron

q=1.6\times 10^{-19}\ C

r= 2 nm

We know that force between two charge given

F=K\dfrac{Q_1Q_2}{r^2}

Now by putting the value

F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}

F=5.67\times 10^{-11}\ N

We know that mass of electron

The mass of electron

m=9.1\times 10^{-31}\ kg

F= m a

a= Acceleration of electron

a= F/m

a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2

a=6.2\times 10^{19} m/s^2

S=ut+\dfrac{1}{2}at^2

initial velocity given that zero ,u=0

20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2

t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}

t=2.5\times 10^{-14}\ s

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