All categories

2 years ago

Which statement best describes particles that are affected by the strong

Physics

2 answers:

Eddi Din [679]2 years ago

4 0

Answer: I think it’s D.

Explanation:

uysha [10]2 years ago

4 0

Answer:

C. Particles that make up a nucleus

Explanation:

A P E X

You might be interested in

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

3 years ago

The electric current running through the wire coil in an electric motor exerts force directly onto

Sveta_85 [38]

C) A powerful magnet

7 0

3 years ago

Read 2 more answers

Normal body temperature. the average normal body temperature measured in the mouth is 310 k . what would celsius and fahrenheit

faltersainse [42]

The question is simply asking us to convert one unit, in this case temperature, to another unit. To do this, we need conversion factors to multiply, divide or relate to the original measurement. We do as follows:

Celsius = Kelvin - 273.15
310 - 273.15 = 36.85 degrees celsius

Fahrenheit = (°C × 9/5) + 32
 (36.85 × 9/5) + 32 = 98.33 degrees fahrenheit

Hope this helps.

7 0

3 years ago

When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their

AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

$F = qvB$

Now this causes a centripetal acceleration

$F = = mv^2/r$

$qvB = mv^2/r$ ...........(i)

$B = mv/(rq)$ ...............(ii)

If accelerating potential V is same and then kinetic energy equals the potential energy difference

$\frac{1}{2} mv^2 = Vq$

$v = \sqrt{(2Vq/m)}$ put these value in equation (ii)

$B = m\frac{\sqrt{(2Vq/m)} }{rq}$

simplifying we get

$B =m \frac{(\sqrt{ 2Vm/q})}{r}$

for same location r will be same in both case

$B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}$ ..............(iii)

$B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}$ ..........(iv)

dividing (iv) and (iii) equation we get

$\frac{B_{10}}{B_{7}} = \sqrt{\frac{m_{10}}{{m_7}} }$

${B_{10}} = B_{7} \sqrt{\frac{m_{10}}{{m_7}} }$

$B_{10} = 0.2574T\sqrt{\frac{ (1.663x10^-26}{(1.165x10^-26)}$

so on solving we get

=0.3075 T

5 0

3 years ago

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

11111nata11111 [884]

Answer:

(a) $Q = 7.28\times 10^{14}$

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

$\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}$

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

5 0

3 years ago