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aleksley [76]
3 years ago
10

Will give brainliest! how does an engineer use physical science?

Physics
1 answer:
pentagon [3]3 years ago
3 0

Answer: gravity, circuits

Explanation:

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Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one en
Airida [17]

Answer: 3 m.

Explanation:

Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by  gravity acting on both children  must be 0.

As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.

If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):

mJill* 5m -mJack* d = 0

60 kg*5 m -100 kg* d =0

Solving for d:

d = 3 m.

6 0
3 years ago
If a person is walking at 1.2 m/s and 60 seconds later the person is running at 10 m/s, what was the acceleration rate?​
Marina CMI [18]
The acceleration rate would be .14667 m/s^2
6 0
3 years ago
Andrea was watching her brother in the ocean and noticed that the waves were coming into the beach at a frequency of 0.777778 Hz
Stolb23 [73]
8 wave units I guess I tried it should be the answer though
3 0
3 years ago
A proton initially has v=4.0i^−2.0j^+3.0k^ and then 4.0 s later has v=−2.0i^−2.0j^+5.0k^ (in meters per second). For that 4.0 s,
pishuonlain [190]

Answer:

a_{avg}=-1.5i+0.5k

1.58113883008\ m/s^2

-18.43^{\circ}\ or\ 161.57^{\circ}

Explanation:

u = 4.0i−2.0j+3.0k v = −2.0i−2.0j+5.0k

Average acceleration is given by

a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{-2-4, -2+2, 5-3}{4}\\\Rightarrow a=-1.5i-0j+0.5k

a_{avg}=-1.5i+0.5k

The magnitude is

a_{avg}=\sqrt{(-1.5)^2+0.5^2}\\\Rightarrow a_{avg}=1.58113883008\ m/s^2

The magnitude is 1.58113883008\ m/s^2

The angle is

\theta=tan^{-1}\dfrac{a_z}{a_x}\\\Rightarrow \theta=tan^{-1}\dfrac{0.5}{-1.5}\\\Rightarrow \theta=-18.43^{\circ}\ or\ 161.57^{\circ}

The angle between a_{avg} and the positive direction of the x axis is -18.43^{\circ}\ or\ 161.57^{\circ}

4 0
3 years ago
Los músicos de una sinfónica siempre ""calientan"" sus instrumentos de viento soplándolos antes de un concierto. ¿Para qué sirve
Snowcat [4.5K]

El calentamiento del instrumento viento antes de un concierto asegura que todos los instrumentos estén afinados y existan un mejor sonido, debido al cambio de frecuencia del instrumentos por los cambios de velocidad del aire debido al cambio de temperatura

Los instrumentos musicales trabajan por procesos de resonancia, ya sean de cuerda o viento; en los instrumentos de viento la frecuencia que emites esta dada por la relación  

                   fₙ = n \ \frac{v_s}{2L}    n = 1, 2, 3, ...

Donde f es la frecuencia emitida, L la longitud del tubo n es una constante entera y v_s es la velocidad del sonido

La velocidad del sonido en el aire depende de la temperatura del aire, según la relacion          

               vs = vo + 0,6 T  

Donde v₀ es la temperatura del aire a 0ºC, v₀ = 331 m/s y T la temperatura en grados centígrados.

De esta dos expresiones podemos ver que la frecuencia que emite el instrumento de viento depende  de la temperatura del aire, además en los instrumentos con boquilla la frecuencia de resonancia de la boquilla también depende de la temperatura de la boquilla que por ser liviana cambia fácilmente.

En conclusión el calentamiento del instrumento antes de un concierto asegura que todos los instrumentos estén afinados y existan un mejor sonido, debido al cambio de frecuencia del instrumentos por los cambios de velocidad del aire debido al cambio de temperatura

aprende mas acerca velocidad sonido aquí:  

brainly.com/question/22160656

8 0
3 years ago
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