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ivann1987 [24]
3 years ago
5

1. If we want to increase the strength of an electromagnet, which 2 of

Physics
1 answer:
kap26 [50]3 years ago
4 0

Answer:

A) Have more coils of wire

F) Use a more powerful battery

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alekssr [168]

Answer:

the answer to the first one is B and the second one is c I hope I got them right

8 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
You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceler
Vanyuwa [196]

Answer:

a) t= 4.81 s,and t= 23.51.

b)d=6.88 m

c)v=4.8 m/s

Explanation:

Acceleration of train ,a= 0.6 m/s²

u = 0 m/s

Your speed ,V= 8.5 m/s

Lets take after t time he you catch the train

Distance travel by train in t time

d=ut+\dfrac{1}{2}at^2

d=\dfrac{1}{2}\times 0.6\times t^2  ----------1

d= V ( t- 4)

d= 8.5 ( t- 4)  --------2

By equating equation 1 and 2

0.3 t² =  8.5 ( t- 4)

0.3 t² -8.5 t + 34 = 0

t= 4.81 s and t= 23.51

It means that you will catch train after t= 4.81 s,and t= 23.51.

t=23.51 sec means that you will catch the train after 23.51 sec also because acceleration of train is low.

Distance travel

d= 8.5 ( t- 4)

t= 4.81 s

d= 8.5 ( 4.81- 4)  m

d=6.88 m

Lets speed = v

0.3 t² =  v ( t- 4)

0.3 t² - v t + 4 v = 0

To have one solution

D=\sqrt{b^2-4ac}

D= 0 Should be zero.

v²- 4 x 0.3 x 4 v

v = 4 x 0.3 x 4

v=4.8 m/s

6 0
3 years ago
Suponga que usted pone en contacto 2 cuerpos, uno con carga +6 y otro con carga -8. ¿Con qué carga queda cada uno al final?
soldi70 [24.7K]
What is the question?
4 0
3 years ago
2 eggs cook in a frying pan, which is sitting over a gas burner. Is thermal energy being transferred from the metal pan to the e
sesenic [268]
Hope that helps have a good day though.

7 0
3 years ago
Read 2 more answers
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