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ivolga24 [154]
3 years ago
10

Sound travels through air at 343 m/s at 20 °C. A bat emits an ultrasonic squeak and hears the echo 0.05 second later . How far a

way was the object that reflected it?
Physics
1 answer:
Rashid [163]3 years ago
7 0

Answer:

Distance of the object is 8.6 m

Explanation:

As we know that the speed of sound at t degree C is given as

v = 332 + 0.6 t

here we know that the temperature is

t = 20 degree C

so we have

v = 332 + 0.6(20)

v = 344 m/s

now we know that bat heard the echo of sound in 0.05 s

so the to and fro distance of the object is d + d = 2d

so we have

2 d = v\times t

2d = 344(0.05)

d = 8.6 m

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This equation is known as the ideal gas law, and it can be used to predict the behavior of many gases at relatively low pressure
Masja [62]
The correct answer is 
<span>C) either the pressure of the gas, the volume of the gas, or both, will increase.

In fact, the ideal gas law can be written as
</span>pV=nRT
<span>where 
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas

We can see that if the temperature T increases, then the term on the right in the equation increases, therefore the term on the left should increase as well. In order for this to be possible, at least one between p and V should increase, or also both of them. Therefore, the correct answer is C.</span>
6 0
3 years ago
A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are
zhannawk [14.2K]

Answer:

y = 17.04 ft

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

y = 180 ft

horizontal distance of the tower from mid point is given as

x = \frac{1300}{2} = 650 ft

now from the equation of parabola we have

y = k x^2

180 = k(650^2)

k = 4.26 \times 10^{-4}

now we have

y = (4.26 \times 10^{-4})x^2

now we need to find the height at distance of 200 ft from center

so we have

y = (4.26 \times 10^{-4})(200^2)

y = 17.04 ft

8 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in
lilavasa [31]

Answer:

Explanation:

The  change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume )  --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume  = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process,  Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case  

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

6 0
4 years ago
On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r
arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s

\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m

From Hooke's law

mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0
3 years ago
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