A group of hikers hear an echo 3.3 s after they shout. The temperature is 20◦C.

Physics

1 answer:

romanna [79]3 years ago

4 0

Answer:

560 m

Explanation:

The speed of sound in air is approximately:

v ≈ v₀ + 0.6T

where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.

The speed of sound at 20°C at that altitude is:

v ≈ 327 + 0.6(20)

v ≈ 339 m/s

The sound travels from the hikers to the mountain and back again, so it travels twice the distance.

339 m/s = 2d / 3.3 s

2d = 1118.7 m

d = 559.35 m

Rounding, the mountain is approximately 560 m away.

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3 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$