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scoray [572]
3 years ago
11

Match each method of transferring electric charge with the correct description

Physics
2 answers:
Mazyrski [523]3 years ago
7 0

Answer:

Friction-1

Conduction-2

Induction-3

Explanation:

Just trust me, bro

babunello [35]3 years ago
4 0

Answer:

FRICTION  OPT 2

INDUCTION OPT 3

CONDUCTION OPT 1

Explanation:

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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
The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p
Blizzard [7]

Answer:

True

Explanation:

In this particular case, the area of the graph represents the impulse.

In fact, impulse is defined as the change in momentum of an object:

I=\Delta p

Moreover, impulse is also defined as the product between the magnitude of the force acting on an object and the duration of the collision:

I=F\Delta t

If we plot a graph of the force versus the time, if the force is constant then this graph will have a rectangular shape, and the area under the graph will simply be the product

F\cdot \Delta t

which corresponds to the definition of impulse.

8 0
3 years ago
Earlier we discussed the concept of isostasy, where lower density rocks rise higher than higher density rocks. How is the variat
Elena L [17]

The variation of water depth at spreading centers (ridges) controlled by isostasy  as convective cooling cools the rocks much more effectively the than heat conduction.

<h3>What is convective heat transfer?</h3>

When heat transfer takes place between the two fluids in direct or indirect contact.

The lithosphere cools when it moves away from the ridge axis by sea floor spreading. The cooler rocks have low density, so the sea floor gets deeper as the lithosphere gets more dense.

Thus, the convective cooling cools the rocks much more effectively the than heat conduction.

Learn more about convective heat transfer

brainly.com/question/10219972

#SPJ1

4 0
2 years ago
What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300
Vinvika [58]
The efficiency of an ideal Carnot heat engine can be written as:
\eta = 1-  \frac{T_{cold}}{T_{hot}}
where
T_{cold} is the temperature of the cold region
T_{hot} is the temperature of the hot region

For the engine in our problem, we have T_{cold}=300 K and T_{hot}=400 K, so the efficiency is
\eta= 1 - \frac{300 K}{400 K}=0.25
4 0
3 years ago
What apparatus can be used to measure the internal diameter of a test tube?​
adelina 88 [10]

Answer:

To determine the volume of a given beaker/calorimeter by measuring its internal diameter and depth with vernier calipers. A vernier caliper is a measuring instrument with two scales: a main scale and a vernier scale that slides over the main scale.

Explanation:

5 0
2 years ago
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