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-BARSIC- [3]
2 years ago
5

A stone is dropped from the

Physics
1 answer:
ICE Princess25 [194]2 years ago
8 0
  • Height=h=500m
  • Acceleration=g=10m/s^2
  • Initial velocity=u=0
  • Speed of sound=c=340m/s
  • TIME TAKEN BY STONE TO HIT WATER=t
  • Time taken by sound to hear back=T

Now

\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2

\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2

\\ \sf\longmapsto 500=5t^2

\\ \sf\longmapsto t^2=100

\\ \sf\longmapsto t=10s

Now

\\ \sf\longmapsto h=cT

\\ \sf\longmapsto T=\dfrac{h}{c}

\\ \sf\longmapsto T=\dfrac{500}{340}

\\ \sf\longmapsto T=1.47\approx 1.5s

Total time:-

\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s

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xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height  = 8.18m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we use one of the motion equations.

            v²  = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

             v² = 0² + (2 x 9.8 x 8.18)

             v² = 160.3

             v = 12.7m/s

7 0
3 years ago
An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo
saveliy_v [14]

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

h = ut + \frac{1}{2}at^{2}

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

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2 years ago
The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

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