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3 years ago

Find the time t it takes the current to reach 99.999% of its maximum value. assume that r=10 ohms and l=50 millihenrys.

1 answer:

6 0

<span>I believe time is equal to r times I which would ten times fifty which would equal 500. I confess to not be an expert on the subject but for some reason T equals R times I came into my head.</span>

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How can two tectonic plates move relative to each other

yarga [219]

They are doing the transform type of tectonic plate movement. This is a process where the plates move sideways. But they move at a rate of a few inches a year. Did that help you?

7 0

3 years ago

A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

faltersainse [42]

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

5 0

3 years ago

Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

5 0

3 years ago

Read 2 more answers

An object initially at rest experiences an acceleration

lidiya [134]

Answer:

1.4m/s

Explanation:

Average velocity is the total distance covered divided by the total time taken.

Average velocity = $\frac{total distance }{time }$

Total time taken = 5s + 6s = 11s

The first distance covered = velocity x time = 1.4 x 5 = 7m

second distance covered = velocity x time = 1.4 x 6 = 8.4m

So;

Average velocity = $\frac{7 + 8.4}{11}$ = 1.4m/s

5 0

3 years ago

At what height is an object that weighs 490 newtons if its gravitational potential energy is 4900 J?

Sergio [31]

Answer:

The height at which the object is moved is 10 meters.

Explanation:

Given that,

Force acting on the object, W = F = 490 N

The gravitational potential energy, P = 4900 J

We need to find the height at which the object is moved. We know that the gravitational potential energy is possessed due to its position. It is given by :

$P=mgh\\\\h=\dfrac{P}{mg}\\\\h=\dfrac{4900\ J}{490\ N}\\\\h=10\ m$

So, the height at which the object is moved is 10 meters. Hence, this is the required solution.

5 0

3 years ago