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andrey2020 [161]
4 years ago
12

The speed of a note created by a flute will increase as the speed of sound increases. When a marching band goes outside on a col

d day, what would you predict would happen to the note played on a flute?
A) It would decrease because the speed of sound and temperature are proportional.
B) It would increase because the speed of sound and temperature are proportional.
C) It would decrease because the speed of sound and temperature are inversely proportional.
D) It would increase because the speed of sound and temperature are inversely proportional.
Physics
2 answers:
krek1111 [17]4 years ago
8 0

Answer:

a

Explanation:

It would decrease because the speed of sound and temperature are proportional. Sound will travel faster when the temperature increases.

Kisachek [45]4 years ago
7 0

The answer is b because im a teacher at apmcota (arthur polly mays

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A 52 kg person is running 2.5 m/s. What is their momentum?
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Answer:

130kg/ms

Explanation:

  1. given data
  2. mass,52kg
  3. velocity,2.5m/s
  • momentum?
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6 0
3 years ago
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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
4 years ago
Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each mo
postnew [5]

Answer:

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

Explanation:

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?

from the principle of momentum

In a system of colliding bodies,we know that the total momentum before collision will equal to the total momentum after collision.

Take note that momentum is the product of mass and velocity

momentum before collision=momentum after collision

mass, m

u=initial velocity of the identical masses

v2=the common velocity after the collision

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ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

8 0
3 years ago
The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist
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Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is  g = 167.2 \ m/s^2  

Explanation:

From the question we are told that

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       The time take is  t =  2.00 *10^{2 \ }s

Generally the period of this oscillation is mathematically evaluated as

         T = \frac{N}{t }

substituting values

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The period of this oscillation is mathematically represented  as

               T = 2 \pi \sqrt{\frac{l}{g} }

making g the subject of the formula we have

              g = \frac{L}{[\frac{T}{2 \pi } ]^2 }

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Substituting values

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               g = \frac{4 * 3.142 ^2  * 1.08 }{0.505^2 }  

              g = 167.2 \ m/s^2  

7 0
3 years ago
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Nonamiya [84]

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8 0
3 years ago
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