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2 years ago

A plane flying against a jet stream will travel faster than a plane traveling with a jet stream.

2 answers:

7 0

Their "airspeeds" (speed through the air) are equal, but the one traveling in the
same direction as the jet-stream appears to move along the ground faster.

koban [17]2 years ago

3 0

<h2><u><em>FALSE, THE PLANE HAS MORE RESISTANCE THAN THE PLANE TRAVELING WITH THE JET STREAM</em></u></h2>

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In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second

wolverine [178]

Answer:

$f_{beat} = 1.64\ Hz$

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

$f' = f(\dfrac{v}{v-v_s})$

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

$f' = 64\times (\dfrac{340}{340 - 8.50})$

f' = 65.64 Hz

now, beat frequency is equal to

$f_{beat} = f' - f$

$f_{beat} = 65.64 - 64$

$f_{beat} = 1.64\ Hz$

hence, beat frequency is equal to 1.64 Hz

3 0

2 years ago

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0

2 years ago

What is physics? In your own words too

denis23 [38]

The study of how the world works

4 0

2 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

5 0

2 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

2 years ago