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4 years ago

How far away is Pluto?

Physics

2 answers:

fgiga [73]4 years ago

4 0

4.7 billion miles or 7.5 billion km far away from Earth.

GaryK [48]4 years ago

3 0

Pluto is 4.67 billion miles away

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What is the magnetic field strength, in units of microTesla (T), at a distance of 6.3 cm from a long straight wire carrying a cu

bezimeni [28]

Answer:

$2.1 \mu T$

Explanation:

The magnetic field strength is given by:

$B=\frac{\mu_0 I}{2 \pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the distance from the wire

In this problem,

I = 6.5 A

r = 6.3 cm = 0.063 m

So, the magnetic field strength is

$B=\frac{(1.256\cdot 10^{-7} H/m)(6.5 A)}{2 \pi (0.063 m)}=2.1\cdot 10^{-6} T=2.1 \mu T$

7 0

3 years ago

If you measured 25 g jelly beans and 25 g of gumdrops, the number of jellybeans would be

KengaRu [80]

The number of jellybeans would be B.more than the gumdrops.

Please rate brainliest.

7 0

3 years ago

Read 2 more answers

How many planets are there?

larisa86 [58]

Their are 8 planets.
<span>Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune.
</span>Their were 9 planets.

4 0

4 years ago

Read 2 more answers

What feature of a DMM allows you to connect the test probes in the circuit, remove the probes, and then read the measured value

Elena L [17]

B. 24 ohms. D. Touch/hold feature

D. From negative to positive =negative charges

B. 24 ohms.

in parallel, combined resistance is less than the lowest resistance

4 0

4 years ago

you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh

inysia [295]

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

$sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}$

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

$d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km$

7 0

4 years ago