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andriy [413]
2 years ago
12

A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi

mately how deep is the well?
Physics
1 answer:
Temka [501]2 years ago
4 0

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

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A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
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so W=12*100
W=1200J
4 0
2 years ago
Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t
Zepler [3.9K]

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    p_{f} = m₁ v₁ + m₂ v₂

   p₀ =  p_{f}

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

6 0
2 years ago
To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster tha
Daniel [21]

Answer:

Explanation:

If the dragster  attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .

So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .

v = u + a t

v₀ = 0 + a tmax

tmax = v₀ / a

The value of tmax is v₀ / a .

5 0
2 years ago
As new-forming stars grow by gravitationally attracting more material, they become brighter. If a star becomes too bright, the r
ratelena [41]

Answer:

Check the explanation

Explanation:

To tackle situations like the one above, the rate of gravity of that star must be equal to the rate of power output but we don’t have radius of that star. Also temp is not mentioned. And emissivity of star is also not mentioned. So the only possible way is like Einstein mass energy relationship E=mc^2=6.5380e40

power =E/Time so this energy is transferred per sec.

4 0
2 years ago
Read 2 more answers
A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta
-Dominant- [34]

The car would go from  zero to 58.0 mph in 2.6 sec.

Since the force on the car is constant, therefore the acceleration of the car would also be constant.

Now for constant acceleration we can use the equation of motion

Using first equation of motion to calculate the acceleration of the car

v=u+at

29=0+a×1.30       ...... Eq. (1)

Again using the first equation of motion

58=0+a*t             ....... Eq. (2)

Dividing eq. (2) with equation 1

t=2×1.3

t=2.6 sec

7 0
2 years ago
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