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Lemur [1.5K]
3 years ago
14

If you double the voltage in a circuit and reduce the resistance by a factor of four what will happen to the current? A) it will

double B) it will decrease by a factor of 2 C) it will increase by a factor of 8 D) it will decrease by a factor of 8
Physics
1 answer:
KatRina [158]3 years ago
5 0
By Ohm's Law, we can relate current, voltage and resistance. It is expressed as V=IR. That is, there is a direct relationship between voltage and resistance and voltage and current.

V = IR
V1/2V1 = I1R1 / I2R1/4
1/2 = 4I1/I2
I2 = 8I1

Therefore, <span> it will increase by a factor of 8. Hope this answers the question.</span>
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Imagine you had a bar of gold and decided to cut in half. You repeated this
MrMuchimi

Answer:

b, an element

Explanation:

this is because you never combined it with any other element in the periodic table and gold is not a compound or mixture.

7 0
3 years ago
A ball rolls down a hill, starting from rest. How long is it rolling if it accelerates at 3m/s2 and ends with a velocity of 35m/
Kryger [21]
<h2>Answer</h2>

The answer to this question is 11.67 s

<h2>Explanation</h2>

As we know that accelartion is the rate of change of velocity. So, it can be write as

a = (V_f -V_i) /t

where

V_f is the final velocity

V_i is the initial velocity

t is the time

a is the accelartion

as we konw

a = 3 ms^{-2}

t = ?

From rest So,

V_i = 0

V_f = 35 ms^{-1}

Putting values

3 = (35 - 0)/t

3 = 35/t

t = 35/3

t = 11.67 s

So, the right answe is 11.67 s

6 0
3 years ago
If you graph the average velocity (y-axis) vs. the elapsed time (x-axis), what
Aleks [24]

Answer:

a diagonal line on the graph

Explanation:

the slope would be constant at an increasing rate

y=x

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4 0
3 years ago
3.7 kg of a saturated water vapor at 0.4 MPa is isothermally cooled until it is a saturated liquid. Calculate the amount of heat
kirill115 [55]

Answer:

7.894 MJ

Explanation:

Given that:

The mass of the saturated water vapor = 3.7 kg

The pressure of the saturated water vapor = 0.4 MPa

From saturated properties of steam tables when the pressure is at 0.4 Mpa

The enthalpy in (kJ/kg);

hf = 604.7 kJ/kg

hg = 2738.1 kJ/kg

The rejected heat during this process is:

Q = m(hg - hf)

Q = 3.7(2738.1 - 604.7) kJ/kg

Q = 3.7(2133.4) kJ/kg

Q = 7893.58 kJ/kg

Q = 7.894 MJ

5 0
3 years ago
What is the first step in the formation of a protostar?
Fittoniya [83]

Star formation begins in relatively small molecular clouds called dense cores.[7] Each dense core is initially in balance between self-gravity, which tends to compress the object, and both gas pressure and magnetic pressure, which tend to inflate it. As the dense core accrues mass from its larger, surrounding cloud, self-gravity begins to overwhelm pressure, and collapse begins. Theoretical modeling of an idealized spherical cloud initially supported only by gas pressure indicates that the collapse process spreads from the inside toward the outside.[8] Spectroscopic observations of dense cores that do not yet contain stars indicate that contraction indeed occurs. So far, however, the predicted outward spread of the collapse region has not been observed.[9]

The gas that collapses toward the center of the dense core first builds up a low-mass protostar, and then a protoplanetary disk orbiting the object. As the collapse continues, an increasing amount of gas impacts the disk rather than the star, a consequence of angular momentum conservation. Exactly how material in the disk spirals inward onto the protostar is not yet understood, despite a great deal of theoretical effort. This problem is illustrative of the larger issue of accretion disk theory, which plays a role in much of astrophysics.

Regardless of the details, the outer surface of a protostar consists at least partially of shocked gas that has fallen from the inner edge of the disk. The surface is thus very different from the relatively quiescent photosphere of a pre-main sequence or main-sequence star. Within its deep interior, the protostar has lower temperature than an ordinary star. At its center, hydrogen is not yet undergoing nuclear fusion. Theory predicts, however, that the hydrogen isotope deuterium is undergoing fusion, creating helium-3. The heat from this fusion reaction tends to inflate the protostar, and thereby helps determine the size of the youngest observed pre-main-sequence stars.[11]

The energy generated from ordinary stars comes from the nuclear fusion occurring at their centers. Protostars also generate energy, but it comes from the radiation liberated at the shocks on its surface and on the surface of its surrounding disk. The radiation thus created most traverse the interstellar dust in the surrounding dense core. The dust absorbs all impinging photons and reradiates them at longer wavelengths. Consequently, a protostar is not detectable at optical wavelengths, and cannot be placed in the Hertzsprung-Russell diagram, unlike the more evolved pre-main-sequence stars.

The actual radiation emanating from a protostar is predicted to be in the infrared and millimeter regimes. Point-like sources of such long-wavelength radiation are commonly seen in regions that are obscured by molecular clouds. It is commonly believed that those conventionally labeled as Class 0 or Class I sources are protostars.[12][13] However, there is still no definitive evidence for this identification.

4 0
3 years ago
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