Ask question

Ask question

All categories

3 years ago

5

An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm

to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 150 m from where he is standing. You are 127 km from the tree. How long does it take for the sound to reach Clark

1 answer:

Rus_ich [418]3 years ago

6 0

Answer:

t = 0.437 s

Explanation:

The speed of sound is a constant that is worth v = 343 m / s

v = d / t

t = d / v

the time it takes for the sound to reach Clark at d = 150 m is

t = 150/343

t = 0.437 s

This same sound takes much longer to reach you

t₂ = 127 10³/343

t₂ = 370 s

You might be interested in

A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What

Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

$T=\dfrac{1}{f}$

$T=\dfrac{1}{215}$

$T=4.65\times 10^{-3}\ s$

(b) Speed of sound in air, v = 343 m/s

It can be given by :

$v=f\times \lambda$

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{343\ m/s}{215\ Hz}$

$\lambda=1.59\ m$

Hence, this is the required solution.

5 0

3 years ago

what is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3

Oduvanchick [21]

V=m×d
m=v/d
m=113/21.4
m=5.28g

6 0

3 years ago

Read 2 more answers

Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109

r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

$E=\Delta mc^2\\\Delta m=\frac{E}{c^2}$ (1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

$E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J$

Hence, by replacing in the equation (1) you have (c=3*10^{8}m/s)

$\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg$

HOPE THIS HELPS!!

3 0

3 years ago

Read 2 more answers

If 41.5 mol of an ideal gas occupies 86.5 L at 27.00 °C, what is the pressure of the gas?

Dmitry [639]

PV=nRT
(P)(86.5)=(41.5)(.08206)(300.15)
(P)(86.5)=(1022.157824)
P=11.81685345 atm

7 0

3 years ago

Read 2 more answers

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

jeka94

Answer:

$1.8\times 105 N/C$

Explanation:

We are given that

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Using 1m=100 cm

We have to find the electric field strength.

$v^2-u^2=2as$

Using the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Mass of electron,m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Substitute the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

7 0

3 years ago