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3 years ago

How many amps are there in the current of a circuit if the voltage source is 140v and the resistance is 2 ohms?

Physics

1 answer:

VLD [36.1K]3 years ago

7 0

Voltage (V)= Current (I) * Resistance (R)
I=V/R=140/2=70A

Hope this helps!

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago

A geosynchronous satellite moves in a circular orbit around the Earth and completes one circle in the same time T during which t

andreyandreev [35.5K]

Answer:

Explanation:

The time period of geosynchronous satellite must be equal to T .

The radius of its orbit will be ( R+ h )

orbital velocity V₀ = $\sqrt{\frac{GM}{( R+h)} }$

Time period T = 2π( R + h ) / V₀

= 2π( R + h ) x $\sqrt{\frac{( R+h)}{GM } }$

$\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ = R +h

h = $\frac{T^\frac{2}{3}(GM)^\frac{1}{3} }{(2\pi )^\frac{2}{3} }$ - R.

7 0

3 years ago

Read 2 more answers

How can you be both at rest and also moving at 100,000 km/h at the same time

4vir4ik [10]

You could be lying completley still on your bed, and all though it seems you are at rest, you are moving along with the earth around the sun and hence are motion. This is why 'being at rest' is more of a relative term. Hope this helps!

7 0

3 years ago

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

8 0

2 years ago

When an exothermic reaction releases thermal energy, this energy is usually_____.

saw5 [17]

Explanation:

A chemical reaction in which heat or energy is released is known as an exothermic reaction.

On the other hand, when two objects are placed together and heat flows from hotter object to colder object then this process is known as conduction. Therefore, energy is dissipated in conduction process.

Since energy released released goes into the atmosphere and is not used anywhere.

Thus, we can conclude that when an exothermic reaction releases thermal energy, this energy is usually not useable to do work and it is dissipated by conduction.

8 0

3 years ago