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2 years ago

If a clock expends 2 W of power from a 1.5 V battery what amount of current is supplying the clock?

1 answer:

7 0

1.3 A
If a clock expends 2 W of power from a 1.5 V battery, what amount of current is supplying
the clock?
solution
as we know
p=vi
i=p/v
=2/1.5
=1.3A

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A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m

katovenus [111]

Answer:

217.43298 m/s

Explanation:

$m_1$ = Mass of bullet = 19 g

$m_2$ = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

$\theta$ = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s$

As the momentum is conserved

$m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s$

The speed of the bullet is 217.43298 m/s

5 0

3 years ago

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'

Mars2501 [29]

I belive its like 1200 mile per hour ive done he math for it

3 0

3 years ago

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 : $x =16R/15$

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2 years ago

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

svetlana [45]

Let's call $m=565~g=0.565~kg$ the mass of the glider and $m_w=7\cdot12~g =84~g=0.084~kg$ the total mass of the seven washers hanging from the string.
The net force on the system is given by the weight of the hanging washers:
$F_{net} = m_w g$
For Newton's second law, this net force is equal to the product between the total mass of the system (which is $m+m_w$ ) and the acceleration a:
$F_{net}=(m+m_w)a$
So, if we equalize the two equations, we get
$m_w g = (m+m_w)a$
and from this we can find the acceleration:
$a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2$

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3 years ago

You’ve made the hypothesis that the stepper the slope , the faster a ball will be rolling when it reaches the bottom .

BigorU [14]

Answer:

B) how steep the slope is

Explanation:

Because you have to know how is the influence of the steep of the slope in the time that a ball reaches the bottom. The steep of the slope is the variable that you would have to change in an experiment.

I hope this is useful for you

regards

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2 years ago