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Nat2105 [25]
2 years ago
12

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

"splat" 2.5s later. How high is the building? The speed of sound is 340m/s. You may ignore air resistance.
Physics
1 answer:
tatiyna2 years ago
7 0

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2

For the sound

Distance = Speed × Time

\text{Distance}=340\times (2.5-t)

Here, distance traveled by the stone and sound is equal

\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0

t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds

The time taken to fall down is 2.4158 seconds

h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m

Height of the buidling is 28.6260196842 m

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grigory [225]

The cars will have equal speeds and the 2 kg car will have greater kinetic energy.

7 0
1 year ago
A submarine completed a 450 km training with an average speed of 50 km/h. For the first 180 km, it travelled at an average speed
Kryger [21]

Answer:

45km/hr

Explanation:

Total distance=450km

Total speed=50km/hr

Total time= distance/speed

=450/50

=9hrs

distance a=180km

speed a=60km/hr

Time a=180/60

=3hrs

Distance b=450-180=270km

Speed b=?

Time b=270/speed b

Total time=time a + time b

9=3+(270/speed b)

270/speed b =9-3

270/speed b =6

6*speed b =270

Speed b=270/6

Speed b=45km/hr

4 0
2 years ago
A uniform piece of wire, 20 cm long, is bent in a right angle in the center to give it an L-shape. How far from the bend is the
zlopas [31]

Answer:

the center of mass is 7.07 cm apart from the bend

Explanation:

the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is

x₁ = L/2 = 20 cm /2 = 10 cm

when the wire is bent in a right angle the coordinates of the new centre of mass will be

x₂ = L₂/2

y₂=  L₂/2

where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2

x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm

y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm

x₂=y₂=X

locating the bend in the origin (0,0) the distance to the centre of mass is

d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm

d = 7.07 cm

5 0
2 years ago
Read 2 more answers
Whennes
rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

v=v_{0}+g*t

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

v=v_{o}+g*t

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

y =   0.5*g*t^{2}

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]

4 0
2 years ago
The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i
faltersainse [42]

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

 

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

 

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

5 0
2 years ago
Read 2 more answers
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