What is the distance a bullet will travel when launched at 166 meters per second for 5.75seconds?

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

$d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m$

Next, you calculate the wavelength of the sound waves by using the following formula:

$\lambda=\frac{v_s}{f}$

vs: speed of sound = 343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

$\lambda=\frac{343m/s}{400Hz}=0.8575m$

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

$\Delta d=3.827m-2.10m=1.727m$

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

$\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2$

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0

3 years ago

A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

GrogVix [38]

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction ( $\mu$ )=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

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

$acceleration=\mu g=0.8\times 9.8=7.84 m/s^2$

s=2.4 m

$0-(u)^2=2\times (-7.84)\times 2.4$

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

$920\times v=(920+2300)\times 6.13$ (conserving momentum)

v=21.45 m/s

7 0

3 years ago

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr

xxMikexx [17]

Answer:

Radius, $r=2.14\times 10^{-5}\ m$

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, $E=4.9\times 10^{-19}\ J$

Kinetic energy is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}$

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

$qvB\ sin90=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }$

$r=2.14\times 10^{-5}\ m$

So, the radius of the circular path is $2.14\times 10^{-5}\ m$ . Hence, this is the required solution.

3 0

3 years ago

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What kind of waves are lasers worked by?

Andrew [12]

A laser is worked by electromagnetic radiation. The term "laser" originated as an acronym for "light amplification by stimulated emission of radiation".

4 0

3 years ago

Which group in the periodic table is known as salt farmers?

adell [148]

<span>Which group in the periodic table is known as salt formers?

The correct option is the last one: Halogen family.
</span><span>
You can find the halogen or "</span>salt formers" in the group 17 of the periodic table. These are:

- Fluorine.
-Chlorine.
- Bromine.
- Iodine.
- Astatine.

All of them are non-metallic elements and they have 7 electrons.

5 0

3 years ago

Read 2 more answers