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alex41 [277]
2 years ago
8

How does the digestive system help the muscular system?

Physics
1 answer:
Sveta_85 [38]2 years ago
8 0

Answer:

I would say the answer is A... but I'm not so sure ....

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According to Newton’s first law of motion what will an object in motion do when no external force acts on it?
PIT_PIT [208]
The object will continue moving in a straight line at constant speed.
4 0
3 years ago
A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel
larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\      (1)

R1 =  13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}            (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

R_{eq}=\frac{23V}{4A}=5.75\Omega

Now, you can calculate the unknown resistor R2 by using the equation (1):

\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega

hence, the resistance of the unknown resistor is 10.31Ω

8 0
3 years ago
Read 2 more answers
a water-balloon launcher with mass 4 kg fires a 0.5 kg balloon with a velocity of 3 m/s to the east. what is the recoil velocity
kotykmax [81]
I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 =  4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s

Therefore; 4 V1 = 0.5 × 3
                   4V1= 1.5
                     V1= 1.5/4
                          = 0.375 m/s










5 0
3 years ago
Read 2 more answers
A new band sensation is playing a concert and recording it for a live album to be released this summer. The band asks the sound
Amanda [17]

The sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

<h3>Amplitude of sound wave</h3>

The amplitude of a sound wave is the maximum vertical displacement of the sound wave.

The sound mixer will need to increase the amplitude of the sound wave produced by the singer.

The increase in the amplitude of the sound wave produced by the lower tune singer will result in increased loudness of the sound.

Thus, the sound mixer will need to increase the amplitude of the sound wave produced by the singer which will increase the loudness of the sound.

Learn more about sound waves here: brainly.com/question/1199084

8 0
1 year ago
Help with this physics task pls
cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

7 0
3 years ago
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