Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.
Answer:
C2H6 up the road to be with its own in
1). The equation is: (speed) = (frequency) x (wavelength)
Speed = (256 Hz) x (1.3 m) = 332.8 meters per second
2). If the instrument is played louder, the amplitude of the waves increases.
On the oscilloscope, they would appear larger from top to bottom, but the
horizontal size of each wave doesn't change.
If the instrument is played at a higher pitch, then the waves become shorter,
because 'pitch' is directly related to the frequency of the waves, and higher
pitch means higher frequency and more waves in any period of time.
If the instrument plays louder and at higher pitch, the waves on the scope
become taller and there are more of them across the screen.
3). The equation is: Frequency = (speed) / (wavelength)
(Notice that this is exactly the same as the equation up above in question #1,
only with each side of that one divided by 'wavelength'.)
Frequency = 300,000,000 meters per second / 1,500 meters = 200,000 per second.
That's ' 200 k Hz ' .
Note:
I didn't think anybody broadcasts at 200 kHz, so I looked up BBC Radio 4
on-line, and I was surprised. They broadcast on several different frequencies,
and one of them is 198 kHz !
Fundamental frequency,
f=v2l=T/μ−−−−√2l
=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√
=58.96Hz
Let, n th harmonic is the hightest frequency, then
(58.93)n = 20000
∴N=339.38
Hence, 339 is the highest frequency.
∴fmax=(339)(58.93)Hz=19977Hz.
<h3>
What is frequency?</h3>
In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.
learn more about frequency refer:
brainly.com/question/254161
#SPJ4
This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.
c = 3 x 10⁸ m/s
E = m c²
1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²
Divide each side by (3 x 10⁸ m/s)²:
Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)
= (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)
= 1.144 x 10⁻³⁰ kg . (choice-1)
This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________
Something like this could have been much more impressive:
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?
Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)
= (2,242 x 10⁶ x 86,400 x 365) joules
= 7.0704 x 10¹⁶ joules .
How much converted mass is that ?
E = m c²
Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s
Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)
= 0.786 kilogram ! ! !
THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !
</span>
