All categories

4 years ago

What is an example of a voltage source

Physics

2 answers:

maxonik [38]4 years ago

8 0

Answer:A generator or a solar panel are 2 examples

melomori [17]4 years ago

8 0

Answer:

i sent attachments

Explanation:

Most sources of electrical energy (the mains, a battery) are modeled as voltage sources. An ideal voltage source provides no energy when it is loaded by an open circuit (i.e. an infinite impedance), but approaches infinite energy and current when the load resistance approaches zero (a short circuit).

You might be interested in

a 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Find the total momentum of the system.

Hunter-Best [27]

Explanation:It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

А masd<br>Of 500kg a raised to a height of 6m In 30s<br>Find (a) Workdone .

Mazyrski [523]

Answer:

Work done is 882000joule.

power is 29400watt.

Explanation:

given,

Mass(m)=500kg

Acceleration due to gravity(g)=9.8m/s²

Height(h)=6m

Time taken(t)=30s

Workdone=?

Power=?

now,

workdone=force*displaxement

= m*g*h

=500*9.8*6

=8,82,000joule

so, the work done by the man is 8,82,000joule.

then,

power=workdone/time taken

=8,82,000/30

=29,400watt

so, the required power to lift a load is 29,400watt.

5 0

3 years ago

in December which will have a longer day length a country im notheren hemisphere or southern hemisphere

astra-53 [7]

Answer:

Southern Hemisphere

Explanation:

3 0

3 years ago

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

3 0

3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

lyudmila [28]

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration $a_c$ , so:

$a_c = w^2r$

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

$a_c$ = 2.6m/s2

So, Replacing the data, and solving for w, we get:

$2.6m/s = w^2(60.5m)$

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2 $\pi$ )= 1.976 rpm

5 0

3 years ago