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inysia [295]
3 years ago
6

Hydrogen fuel cells are used on the space shuttle to provide the shuttle with all of its electrical energy. Explain why fuel cel

ls are used instead of batteries.
Physics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

Hydrogen used in fuel cells has the energy to weight ratio ten times greater than lithium-ion batteries. Consequently, it offers much greater range while being lighter and occupying smaller volumes. It can also be recharged in a few minutes, similarly to gasoline vehicles.

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If 2 electrically charged objects repel one another , which of the statements could be true, one object has more electrons, the
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Both objects have the same electrical charge. Opposite charges attract. And if they were neutral they would not do anything.
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3 years ago
Read 2 more answers
Help me answer this ASAP
olga_2 [115]

Answer:

The correct answer is Option A.

7 0
3 years ago
A large fake cookie sliding on a horizontal surface isattached to one end of a horizontal spring with spring constantk = 425 N/m
omeli [17]

Answer:

Explanation:

spring constant k = 425 N/m

a ) At the point of equilibrium

restoring force = frictional force

= kx = 10 N

425 x = 10

x = 2.35 cm

b )

Work done by frictional force

= -10 x 2.35 x 10⁻² x 2  J ( Distance is twice of 2.35 cm )

= - 0.47 J

= Kinetic  energy remaining with the cookie  as it slides back through the position where the spring is unstretched .

= 425 - 0.47

= 424.53 J

=

4 0
3 years ago
On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi
Lostsunrise [7]

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

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3 years ago
A small spotlight mounted in the bottom of a swimming pool that is 5.0 m deep emits light in all directions. On the surface of t
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The radius is higher than the refraction
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