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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Which of the following best defines spring constant ?

katovenus [111]

Answer:

Explanation:

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3 years ago

Write an expression for the potential energy UD of a particle when it is at a distance D from the force center, in terms of B an

nadezda [96]

Answer:

E+mc+UD-3

Explanation:

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4 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

Read 2 more answers

A 4 kg block is pushed up an incline that makes a 30° angle with the horizontal, as shown in the figure. Once the block is pushe

Sophie [7]

Answer:

B) 100 J

Explanation:

Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...

∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J

8 0

3 years ago

Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol

valentina_108 [34]

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

$tan 16^0 = \dfrac{h}{x}$

$x =\dfrac{h}{tan 16^0}$

now,

again applying

$tan 20^0 = \dfrac{h}{300-x}$

$300-x=\dfrac{h}{tan 20^0}$

$300-3.49 h=2.75 h$

$6.24h = 300$

h = 48.077 ft

3 0

3 years ago