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3 years ago

Which statement best describes the difference between strong nuclear forces and weak nuclear forces? (2 points)

2 answers:

8 0

Weak nuclear forces are involved when certain types of atoms break down. Strong nuclear forces are responsible for holding atoms' nucleus together. Both forces are actually very strong, and are only felt at very short distances. They are 2 of the 4 fundamental forces that govern all matter.

san4es73 [151]3 years ago

8 0

Answer:

Weak nuclear forces are involved when certain types of atoms break down. Strong nuclear forces are responsible for holding atoms' nucleus together

Explanation:

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Read 2 more answers

If two point masses 1kg & 4kg are seperated by a distance of 2m. Magnitude of gravitational force exerted by 1kg on 4kg is ?

Aliun [14]

Answer:

F = G Newtons

Explanation:

Given:

Mass of 1st body = $1\:kg$
Mass of 2nd body = $4\:kg$

To Find:

Magnitude of gravitational force

Solution:

Here, we have a formula

$F=\dfrac{G.M_{1}.M_{2}}{r^{2}}$

Substituting the values

$\implies\:\:F = \dfrac{G(1)(4)}{2^{2}}$

$\implies\:\:F = \dfrac{4G}{4}$

$\implies\:\:F = \dfrac{\cancel{4}G}{\cancel{4}}$

$\implies\:\:\red{F = G}$

Know More:

The applied formula for the above solution is

${\boxed{F_{G}=\dfrac{G.M_{1}.M_{2}}{r^{2}}}}$

where,

F $_{G}$ = Gravitational force
G = Gravitational constant
M $_{1}$ = mass of 1st body
M $_{2}$ = mass of 2nd body
r = distance between two bodies

6 0

3 years ago

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago