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In-s [12.5K]
2 years ago
7

If a ball is dropped off a tall building and accelerates at 9.8 m/s^2 until reaching the ground at a speed of 55 m/s, how long w

ill it take for the ball to reach the ground?
Physics
1 answer:
bearhunter [10]2 years ago
8 0

Answer:

5.61s

Explanation:

\pink{\frak{Given}}\begin{cases} \textsf{ A ball is dropped off a tall building and accelerates at 9.8 m/s$^2$ .}\\\textsf{It reaches the ground at a speed of 55 m/s .} \end{cases}

And we need to find out the time taken by the ball to reach the ground .

  • The initial velocity of the ball will be 0 since it is dropped .

Here we can use the First equation of motion , namely ;

\sf \longrightarrow v = u + at

where the symbols have their usual meaning . Now substituting the respective values , we have,

\sf \longrightarrow 55m/s = 0m/s + 9.8 m/s^2(t)\\

\sf \longrightarrow 55m/s = 9.8m/s^2(t)\\

\sf \longrightarrow t = \dfrac{55m/s}{9.8m/s^2}\\

\sf \longrightarrow \boxed{\bf time = 5.61 s }

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Three resistors of resistances, R1=10Ω, R2=5Ω, R3=20Ω are connected in a circuit in such way that same amount of current flows t
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Explanation:

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RT = R1 + R2 + R3

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5 10 15 RT + R1 + R2 + R3 + +

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Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
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