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Olin [163]
3 years ago
9

Which of the following statements about the nuclear strong force is true?

Physics
1 answer:
lisabon 2012 [21]3 years ago
4 0

Answer: 2, the nuclear strong force drops to practically nothing at large distances.

Explanation: The protons and neutrons in the nucleus share subatomic particles called pions. This exchange is what keeps the protons and neutrons stuck together in the nucleus. Despite the strong force being the strongest force, it has a very small range. This is because pions have very short lifespans. So, the strong force would have literally no effect at large distances.

Hope that helped! :)

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How do t cells differ from b cells
Natalka [10]
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B cells can connect to antigens right on the surface of the invading virus or bacteria. 
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for more info , comment !
5 0
3 years ago
ASAP ONLY CORRECT ANSWERS PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEE I RLLY NEED HELPPPPPPP
nadya68 [22]

Answer:

attached below

Explanation:

5 0
3 years ago
Pls help answer embed <br>​
Savatey [412]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

Weight of the object = drag force

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

5 0
3 years ago
Very wierd question<br><br> How big is a black hole?
n200080 [17]

Answer:

about 23.6 billion miles

Explanation:

because it is lol

4 0
3 years ago
In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini
zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and v_1 , v_2 be the final velocity of 8 ball and cue ball respectively

\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}

The angle after which cue ball is deflected is given by

\theta _1=90-40=50^{\circ}

Conserving momentum in x direction

mu=mv_1cos40+mv_2cos50

3=v_1cos40+v_2cos50

Along Y axis

0+0=v_1sin40-v_2sin50

v_1sin40=v_2sin50

substitute the value of v_1

we get v_2=1.912 m/s

v_1=2.27 m/s

5 0
3 years ago
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