If the frequency of a periodic wave is doubled, the period of the wave will be

Do some research on digital vs. analog recording, and decide which you think gives the most accurate and best quality sound. Sta

rodikova [14]

Answer:

Sound quality depends on a lot of factors, and it is impossible to definitively state that either analog or digital is fundamentally better. These days, many records are made using playback of a digital file, so vinyl preference cannot be attributed solely to the differences in the way the sound wave is reproduced.

Explanation:

6 0

2 years ago

A scientist is planning an experiment that will include the measurement of temperatures. She wishes to avoid using negative valu

cricket20 [7]

Answer:

C.

Explanation:

Fahrenheit scale is a temperature scale that measures the boiling point of water at 212 degrees F and the freezing point at 32 degrees F. The Fahrenheit scale was developed by the German scientist Daniel Gabriel Fahrenheit in 1724.

In the given scenario, the scientist, who wishes to measure the temperature of her experiment which will expectantly be colder than the point of water, should use the Fahrenheit scale.

So, the correct answer is option C.

8 0

2 years ago

A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.4 m/s, a typical raindrop speed. How fast

kenny6666 [7]

The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1$ is the mass of the first mosquito

$u_1 = 0$ is the initial velocity of the mosquito

$m_2 = 50 m_1$ is the mass of the raindrop

$u_2 = 8.4 m/s$ is the initial velocity of the raindrop

$v$ is the final combined velocity of the raindrop+mosquito

Re-arranging the equation and substituting, we find:

$m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s$

Learn more about momentum here:

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#LearnwithBrainly

4 0

3 years ago

In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ

Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

$Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J$

8 0

3 years ago

A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago