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2 years ago

What happens if 5 Protons combine with 4 neutrons.

2 answers:

7 0

in the case of beryllium, a light metal, only the beryllium-9 isotope is stable with its 9 nucleons (i.e. 4 protons and 5 neutrons)

Hope this helps so sorry if it does not help

JulijaS [17]2 years ago

6 0

They go looking for 5 electrons to join them and make one whole atom of Boron.

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Which of the following is not a symptom associated with hypertension

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3 years ago

What are the properties of bungee gum,hmmm?

lys-0071 [83]

It’s both a solid and a liquid. It can thicken and soften depending on how it’s handled. It can be used to cover wounds to stop bleed, and used to drown enemies. Bungee Gum has the properties of both rubber and gum.

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2 years ago

Read 2 more answers

A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with

mariarad [96]

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

$F_friction=u_s*N=u_s*(mg)$

$F-F_friction=ma$

$F-u_s*(mg)=ma$

Second Newton law for the Big Block:

$F_friction=7ma$

$u_s*(mg)=7ma$

$a=u_s*g/7$

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2 years ago

A student designs an electromagnet, as shown in the picture. The electromagnet is only able to pick up 1 paper clip. List 2 modi

Lerok [7]

-- Change the battery to one with higher voltage.

-- Wrap more turns of wire around the spike.

-- If the spike is made of anything else but iron, replace it with a pure iron one.

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2 years ago

A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge

Troyanec [42]

Answer:

$I_{l} =44.84 \mu A$

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire = 1.30×10⁻² Ω

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

$\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))$

$\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{V}{R}$

$I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}$

$I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))$

$I_{l} =44.84 \times 10^{-6}\A$

$I_{l} =44.84 \mu A$

3 0

3 years ago