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olya-2409 [2.1K]

3 years ago

14

Which of the following pairings are more likely to be held together with the strong nuclear force?

1 answer:

aivan3 [116]3 years ago

8 0

Which of the following pairings are more likely to be held together with the strong nuclear force

Explanation:

1.What does a strong nuclear force do in an atom? It repels electrons from other electrons. It repels protons from other protons. It attracts protons and neutrons.

2.The chain reaction requires both the release of neutrons from fissile isotopes undergoing nuclear fission and the subsequent absorption of some of these neutrons in fissile isotopes.

3.The strong nuclear force holds most ordinary matter together because it confines quarks into hadron particles such as the proton and neutron. In addition, the strong force binds these neutrons and protons to create atomic nuclei.

You might be interested in

Which evidence supports the big band theory select 3 options

ruslelena [56]

Answer:

Explanation:

There are different theories and evidence about the big bang, in this case, we're going to see three evidence.

The galaxies are moving from us, this means space is expanded, this in consequence Big Bang's explosion.

The cosmic microwave background radiation is related to the early warmth of the universe.

The observed abundance of hydrogen, helium, deuterium, lithium, these are checked from the spectra of the oldest stars.

5 0

3 years ago

Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com

Tpy6a [65]

M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h = 0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW

5 0

4 years ago

A cart, which has a mass of 2.30 kg is sitting at the top of an inclined plane, which is 4.50 meters long and meets the horizont

expeople1 [14]

Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) The magnitude of the force that causes the cart to roll down is 5.47 N.

c) The acceleration of the cart is 2.38 m/s²

d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

Where:

EP = gravitational potential energy.

m = mass of the object.

g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

4.50 m · sin 14.0° = height

height = 1.09 m

Then, the gravitational potential energy will be:

EP = m · g · h

EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J

The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

Fwx = 5.47 N

The magnitude of the force that causes the cart to roll down is 5.47 N.

c)Using the equation of Fwx we can calculate the acceleration of the cart:

Fwx = m · a

Where "m" is the mass of the cart and "a" is the acceleration.

Fwx / m = a

5.47 N / 2.30 kg = a

a = 2.38 m/s²

The acceleration of the cart is 2.38 m/s²

d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:

x = 1/2 · a · t²

4.50 m = 1/2 · 2.38 m/s² · t²

2 · 4.50 m / 2.38 m/s² = t²

t = 1.94 s

It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:

v = v0 + a · t

v = 0 m/s + 2.38 m/s² · 1.94 s

v = 4.62 m/s

The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the cart.

v = velocity of the cart.

KE = 1/2 · 2.30 kg · (4.62 m/s)²

KE = 24.6 J

The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

The gain of kinetic energy is equal to the loss of gravitational potential energy.

g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:

W = -ΔPE

W = -(final potential energy - initial potential energy)

W = -(0 - 24.6 J)

W = 24.6 J

This can also be calculated using the definition of work:

W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

4 0

4 years ago

A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her

lisabon 2012 [21]

Answer:

10.09 N

Explanation:

Analogously to Newton's second law, torque can be defined as:

$\tau=I\alpha$

Here, I is the moment of inertia and $\alpha$ is the angular acceleration. We have:

$\tau=(0.65kg*m^2)(29.5\frac{rad}{s^2})\\\tau=19.18N*m$

Torque is the vector product of the position vector of the point at which the force is applied by the force vector:

$\vec{\tau}=\vec{r}\times \vec{F}$

Since the effective lever arm is perpendicular to the force, the angle between them is $90^\circ$ . The magnitud of this vector product is defined as:

$\tau=rFsen\theta$ .

Solving for F and replacing the known values:

$F=\frac{\tau}{rsen\theta}\\F=\frac{19.18N*m}{1.9m(sen90^\circ)}\\F=10.09N$

8 0

3 years ago

Two positive charge spheres, the spheres are separated by 0.40 m. The charge on the first sphere is 100 microcoulombs and the

agasfer [191]

Answer: 168.75 N

Explanation:

first, let's convert microcoulombs to coulombs

q1 = 1e-4 C

q2 = 3e-5 C

r = 0.4 m

then use the equation Fe = $\frac{kq_{1} q_{2}}{r^{2} }$

plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2

F = 168.75 N

8 0

1 year ago