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3 years ago

Type of force that holds the nucleus of an atom together

Physics

1 answer:

Kamila [148]3 years ago

7 0

Answer:

Nuclear Forces

Explanation:

The type of force that holds the nucleus of an atom together is called Nuclear Forces.

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How might deforestation lead to desertification

qwelly [4]

If we keep cutting down tree's and if there are enough forest fires then it will turn into a desert....Hope it helps!

6 0

3 years ago

Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite

Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass $_{Hank$ × Acceleration $_{Hank$ = Mass $_{Henry$ × Acceleration $_{Henry$

Mass $_{Hank$ / Mass $_{Henry$ = Acceleration $_{Henry$ / Acceleration $_{Hank$

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass $_{Hank$ / Mass $_{Henry$ = 1 / 1.26

Mass $_{Hank$ / Mass $_{Henry$ = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$