How does the strong nuclear force compare with the electrostatic force in the nucleus of an atom?

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

4 0

3 years ago

Based on what you’ve learned from the raisin cake analogy, which two properties of distant galaxies do astronomers have to measu

Kisachek [45]

Here is the answer to the given question above. Based on the Raisin Cake Analogy, the two properties of distant galaxies that astronomers have to measure to show<span> that we live in an expanding universe are DISTANCE and SPEED. Hope this answers your question. Have a great day!</span>

4 0

3 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

(c) the maximum speed attained by the object during its motion

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

3 years ago

The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.

max2010maxim [7]

Answer:

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg , L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3

α = 1.193 rad / s²

ac = α *2 ⇒ ac = 1.193 rad / s² * 2

ac = 2.86 m / s²

4 0

3 years ago

Marys airplane trip took 5.8 hours for one-half of that time, the airplane flew at a constant speed of 640 miles per hour and fo

Olenka [21]

Distance is speed x time. Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed. This is easy since it was divided in two equal time slices. Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi

7 0

3 years ago