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harkovskaia [24]

2 years ago

14

Johnny was playing baseball with his friends and they noticed a bolt of lightning. They heard thunder seven seconds later. How f

ar away is the storm?

2 answers:

Thepotemich [5.8K]2 years ago

4 0

That particular strike was very roughly 2.4 km (1.5 miles) away from them.

That's if you use 340 m/s (1120 ft/sec) for the speed of sound.
But the air in the region for several thousand feet around a thunderstorm
is doing weird things to sounds that pass through it, so you can't use any
exact number for the speed of sound in a stormy area.

The only thing you can be absolutely sure of is that Johnny and his friends
need to round up their equipment and get in the house. NOW !

murzikaleks [220]2 years ago

3 0

2,317 meters on edg!!

You might be interested in

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

2 years ago

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel

sashaice [31]

1) $3.38\cdot 10^6 m/s$

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

$F_E = F_B\\qE = qvB$

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

$v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s$

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

$qvB=m\frac{v^2}{r}$

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

$r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm$

3) $7.6\cdot 10^{-9}s$

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, $2 \pi r$ , and the period, T:

$v=\frac{2\pi r}{T}$

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

$T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s$

7 0

2 years ago

A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k

9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg : mass of the skier

θ =30° :angle θ of the ski slope with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W= 92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N= 0.10*780.8 N

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax , ax= a : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 = (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0

2 years ago

Have you ever chewed on a wintergreen mint in front of a mirror in the dark? If you have, you may have noticed some sparks of li

lutik1710 [3]

Answer:

Part a)

$E = 3.66 eV$

Part b)

$\lambda = 508.5 nm$

Explanation:

Part a)

change in the energy due to decay of photon is given as

$E = h\nu$

here we know that

$\nu = 8.88 \times 10^{14} Hz$

now we have

$E = (6.6 \times 10^{-34})(8.88 \times 10^{14})$

$E = 5.86 \times 10^{-19} J$

$E = 3.66 eV$

Part b)

While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy

so we have

$\Delta E = \frac{2}{3}(3.66 eV)$

$\Delta E = 2.44 eV$

now to find the wavelength we have

$\Delta E = \frac{hc}{\lambda}$

$2.44 = \frac{1242}{\lambda}$

$\lambda = 508.5 nm$

3 0

2 years ago

The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0

2 years ago