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DIA [1.3K]
3 years ago
13

one thing that stringed, wind, and percussion instruments have in common in regard to the sounds they produce is that a. the ins

truments are struck by tools to produce vibrations and sounds. b. the frequency is always a direct result of adjusting the length of the instrument. c. all of the instruments can be easily adjusted to produce a tone with a certain definite pitch. d. the sound originates from a vibration.
Physics
2 answers:
mario62 [17]3 years ago
3 0

The answer is:

d) the sound originates from a vibration.

The explanation:

The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The sound source creates vibrations in the surrounding medium. As the source continues to vibrate the medium, the vibrations propagate away from the source at the speed of sound, thus forming the sound wave.

LenaWriter [7]3 years ago
3 0
D. The sound originates from a vibration. With winds, it is the vibration of air, strings, the vibration of the string, and percussion, the vibration of various things depending on the instrument. Also, all sounds are produced by vibrations
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What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.
elena-14-01-66 [18.8K]

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

r= \sqrt{\frac{Gm1m2}{F} }

r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m

Answer:

The distance is 0.96m

Explanation:

Given

m1= 900kg

m2= 1600kg

Force F= 0.0001nN

G=6.67430*10^-11 Nm^2/kg^2

Required:

The distance r

Step two:

the formula for the force is given as

F = Gm1m2/r2

make r subject of the formula

r= \sqrt{\frac{Gm1m2}{F} }

r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m

3 0
3 years ago
Read 2 more answers
Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg
nata0808 [166]

Answer:

a) U_{e} = 3 \times 10^{10}\,J, b) v \approx 7745.967\,\frac{m}{s}

Explanation:

a) The potential energy is:

U_{e} = Q \cdot \Delta V

U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)

U_{e} = 3 \times 10^{10}\,J

b) Maximum final speed:

U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }

The final speed is:

v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }

v \approx 7745.967\,\frac{m}{s}

3 0
3 years ago
In a certain region of space, the electric potential is V(x,y,z)=Axy-Bx^2+Cy , where A,B , and C are positive constants.a) Calcu
laiz [17]

Answer:

a)  Eₓ = - A y + 2B x , b)  Ey = -Ax –C , c) Ez = 0 , d) The correct answer is 3

Explanation:

The electric field and the electric power are related

                    E = - dV / ds

a) Let's find the electric field on the x axis

                  Eₓ = - dV / dx

                  dV / dx = A y - B 2x

                  Eₓ = - A y + 2B x

b) calculate the electric field on the y-axis

                Ey = - dV / dy

                dV / dy = A x + C

                Ey = -Ax –C

c) the electric field on the z axis

              dv / dz = 0

              Ez = 0

.d) at which point the electric field is zero

Since the electric field is a vector quantity all components must be zero

X axis

              0 = = - A y + 2B x

              y = 2B / A x

Axis y

             0 = -Ax –C

              .x = -C / A

We substitute this value in the previous equation

             .y = 2B / A (-C / A)

             .y = 2 B C / A2

The correct answer is 3

6 0
3 years ago
Starting from rest, the distance a freely-falling object will fall in 0.50 second is?
FinnZ [79.3K]

Answer:

1.23 m

Explanation:

The vertical distance covered by a free-falling object starting from rest in a time t is

y=\frac{1}{2}gt^2

where

g = 9.8 m/s^2 is the acceleration due to gravity

In this problem, we have

t = 0.50 s

So the distance covered is

y=\frac{1}{2}(9.8 m/s^2)(0.50 s)^2=1.23 m

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kozerog [31]
It's 3600 m and 60 sec
6 0
3 years ago
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