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3 years ago

When current increases and resistance remains constant what happens to voltage?

1 answer:

4 0

According to Ohm's law, V=IR (voltage equals current times resistance). So if thevoltage increases, then the current increases provided that the resistance remains constant

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Bromine, a liquid at room temperature, has a boiling point

lukranit [14]

Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.

3 0

3 years ago

How was the speed of light determined

goblinko [34]

Presently, the speed of light in a vacuum is defined to be exactly 299,792,458 m/s (approximately 186,282 miles per second). . An early experiment to measure the speed of light was conducted by Ole Romer, a Danish physicist, in 1676. Using a telescope, Ole observed the motions of Jupiter and one of its moons, Io

5 0

3 years ago

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The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

NeX [460]

Explanation:

Formula to represent thrust is as follows.

F = $\frac{dP}{dt}$

= $\frac{2p}{dt}$

or, p = $\frac{E}{c}$

$\frac{p}{dt} = \frac{W}{c}$

F = $\frac{2IA}{c}$

= $\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}$

= 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

8 0

3 years ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

7 0

3 years ago

Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n

Brut [27]

The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9 N. m2 / C2

Given are the following:
Q = 1.602 × 10^–19 C
r = 38 x 10^-9 m

Substitue the given:
E = $\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }$

E = 998.476 kN/C

8 0

3 years ago

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