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fredd [130]
3 years ago
5

Beryllium has a nucleus composed of protons and neutrons. Given the data, how many neutrons are in a typical Beryllium nucleus?

Physics
2 answers:
shepuryov [24]3 years ago
6 0
Beryllium has a mass number of 9 and 4 protons.  To find the number of neutrons you need to subtract the number of protons from the mass number.  so...
neutrons = mass - # protons
neutrons = 9 - 4 = 5
Semmy [17]3 years ago
6 0
I believe the answer is C
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bc it was a universal explosion and It started the future

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A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The
Yakvenalex [24]

Answer:

the magnitude of acceleration will be 1.50m/s^2

Explanation:

To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a    from her you can divide both sides by 10 to isolate a:

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a=1.50m/s^2 (which is the typical unit for acceleration)

7 0
3 years ago
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

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v = r\omega

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F_c = m\frac{v^2}{r}=m\omega^2 r

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F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

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\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

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A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



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