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Usimov [2.4K]
3 years ago
8

How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and th

e air temperature is 90.0°F?
Physics
1 answer:
alexandr1967 [171]3 years ago
8 0

Answer:

t = 4.58 s

Explanation:

In this problem, we need to find the time elapse between seeing lightning and hearing the thunder if the lightning strikes 1mi (5280 ft) away and the air temperature is 90.0°F.

T = 90.0°F = 32.2 °C

The speed of sound at temperature T is given by :

v = (331.3 +0.6T)

Put T = 32.2°C

So,

v = (331.3 +0.6(32.2))

= 350.62 m/s

We have, distance, d = 1 mile = 1609.34

So,

t=\dfrac{d}{v}\\\\t=4.58\ s

So, the required time is equal to 4.58 seconds.

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What to do if one day your partner wants you but the next day are unsure about you should I try to make it work or leave ..
olchik [2.2K]
Leave them, listen you are better than being with someone who is unsure. If you are 100% in this they need to match the energy. The purpose of dating is going into marriage eventually and imagine if when you get married they are like “wait I don’t know actually”. True me dump them and don’t look back
6 0
3 years ago
A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole
sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0
3 years ago
At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c
fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

\rho = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

8 0
3 years ago
Parker completed 4 laps around a 400 m track. He ran for a total of 30 mins. What is the
Over [174]

Answer:

Distance: 1600 m Displacement: 0

Explanation:

The distance is because He ran 400 meters 4 times getting 1600 m

4*400=1600

The displacement is 0 because displacement is the total distnce away from the starting point and since he ran laps around the track in the end he ended up in the same spot as last time.

7 0
3 years ago
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