A ball is thrown from the ground with nonzero horizontal and vertical initial velocities. While the ball is in the air, assuming that only the force of gravity acts on it, which of the following statements is true? a. Both the horizontal and vertical components of momentum are constant. b. Only the horizontal component of momentum is constant. c. Neither the horizontal nor the vertical components of momentum are constant.

Answer:

The total momentum is  [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]

Explanation:

The diagram illustration this  system is shown on the first uploaded image (From physics animation)

From the question we are told that

     The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]

      The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]

      The mass of the second object is  [tex]M_2 = 4 \ Dalton[/tex]

      The speed of the second object is  assumed to be  [tex]- v_2[/tex]

The total momentum of the system is the combined momentum of both object which is mathematically represented as

           [tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]

   substituting values

            [tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]

            [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]

Answer:

The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.

Explanation:

The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?

Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = [tex]I[/tex]

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = [tex]5I[/tex]  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

therefore,

for the first case, the K.E. is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

and for the second case, the K.E. is given as

[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]

[tex]KE = \frac{1}{10}Iw^{2}[/tex]

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Now, let's consider moment of inertia =  I  and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia =   (five times larger)

angular speed = ω/5  (five times smaller)

The kinetic energy of a spinning body is given as:

[tex]K.E.=\frac{1}{2} I. w^2[/tex]

On substituting the values, we will get:

[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

This question involves the concept of the law of conservation of momentum.

Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.

The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.

Mathematically,

[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]

where,

m₁ = mass of the first object

m₂ = mass of the second object

u₁ = velocity of the first object before the collision

u₂ = velocity of the second object before the collision

v₁ = velocity of the first object after the collision

v₂ = velocity of the second object after the collision

Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.

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The attached picture illustrates the law of conservation of momentum.

Answer:

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]

The value of the spring constant is 187,500N/m

Answer:

The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.

Explanation:

The period of oscillation is calculated as;

[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]

where;

L is the length of the pendulum bob

g is acceleration due to gravity

If we make L the subject of the formula in the equation above, we will have;

[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]

The length of the pendulum depends on acceleration due to gravity (g).

Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.

Answer:

The tube should be held vertically, perpendicular to the ground.

Explanation:

As the power lines  of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its  potential.

And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.

Answering the two questions in reverse order:

-- No. I don't need to know how the speed of the person changed before I can answer the question.  I can answer it now.

-- The NET work done by the gravitational force is zero.

-- As the person and his girl-friend go up the first half of the wheel, the motor does positive work and gravity does negative work.

-- After they pass the peak at the top and come down the second half of the wheel, the motor does negative work and gravity does positive work, even though the couple may be interested in other things during that time.

-- The total work done by gravity in one complete revolution is zero.

-- The total work done by the motor in one complete revolution is only what it takes to pay back the energy robbed by friction and air resistance.

The work done by the gravitational force is zero.

Work done by a conservative force is path independent. Which means it only depends on the initial and final position of the body. The gravitational force is a conservational force and the gravitational potential energy depends only upon the height of the body.

Let the lowest point of the body is at some height h, then the initial gravitational potential energy of the person is:

PE(initial) = mgh

The final position of the person is also at a height h, thus, the final gravitational potential energy :

PE(final) = mgh

According to the work-energy theorem:

work done = - change in potential energy

work done = -(mgh - mgh) = 0

Thus, the work done is zero in the given case.

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The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Answer:

a.     E = 1.02*10^-27 J

b.     E = 6.39*10^-9eV

Explanation:

a. In order to calculate the energy of the radio photon, you use the following formula:

[tex]E=hf[/tex]             (1)

h: Planck's constant = 6.626*10^-34 Js

f: frequency of the photon = 1545kHz = 1.545*10^6 Hz

Then, by replacing you obtain the energy of the photon:

[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]

b. In electron volts, the energy of the photon is:

[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]

Answer:

Mass of the rope = 2.8 kg

Explanation:

The speed of waves travelling through a rope with linear density (μ) and under tension T is given as v = √(T/μ)

The speed of waves in the rope is also calculated as

v = (d/t)

d = L = length of the rope = 15 m

t = time taken for the wave to move through the rope = 0.545 s

Speed = v = (15/0.545) = 27.523 m/s

Speed = v = √(T/μ)

T = tension in the rope = 140 N

μ = linear density = ?

27.523 = √(140/μ)

27.523² = (140/μ)

(140/μ) = 757.512

μ = (140/757.512) = 0.1848155556 = 0.1848 kg/m

Linear density = μ = (m/L)

m = mass of the rope = ?

L = length of the rope = 15 m

0.1848 = (m/15)

m = 0.1848 × 15 = 2.77 kg = 2.8 kg to 1 d.p.

Hope this Helps!!!

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

The ball reaches the maximum height of 54 feet

The question is about projectile motion,

the ball is shot at an angle α = 45°, and

the initial velocity u = 46 ft/s.

Under the projectile motion, the maximum height H is given by:

[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]

where, g = 9.8 m/s²

substituting the given values we get:

[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]

Hence, the maximum height is 54 feet.

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Answer:

see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Explanation:

This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5

Let's analyze the light beam path emitted by the diver.

* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º

* also the wavelength of light in a material medium changes

      λ_n =  λ / n

where  λ_n is the wavelength in the material and  λ the wavelength in the vacuum air and n the refractive index.

If we include these aspects, the constructive interference equation is

       2t = (m + ½)  λ_n

       2nt = (m + ½)  λ

let's apply this equation to our case

            λ = 2nt / (m + ½)

The incidence of replacement of the oil with respect to water is

        n = n_oil / n_water = 1.5 / 1.33

        n = 1,128

let's calculate

        λ = 2 1,128 t / (m + ½)

        λ = 2,256 t / (m + ½)

In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation

          t = 0.001 mm = 1 10⁻⁶ m

the formula remains

        λ = 2,256 10⁻⁶ / (m + ½)

Let's find what values ​​of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m

     m + ½ = 2,256 10⁻⁶ / λ

     m = 2,256 10⁻⁶ / λ - ½

light purple lan = 400 10⁻⁹m

     m = 2,256 10-6 / 400 10⁻⁹ - ½

     m = 5.64 - 0.5

     m = 5.14

     m = 5

red light  λ = 700 10⁻⁹m

      m = 2,256 1-6 / 700 10⁻⁹ - ½

      m = 3.22 - 0.5

      m = 2.72

      m = 3

we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Answer:

P₃₅₀ = 0.28 watt

Explanation:

First we find the resistance of the wire at 20°C:

R₀ = ρL/A

where,

ρ = resistivity = 1 x 10⁻⁶ Ωm

L = Length of wire = 100 cm = 1 m

A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²

Therefore,

R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)

R₀ = 1.27 Ω

Now, from Ohm's Law:

V = I₀R₀

where,

V = Potential Difference = ?

I₀ = Current Passing at 20°C = 0.5 A

Therefore,

V = (0.5 A)(1.27 Ω)

V = 0.64 volts

Now, we need to find the resistance at 350°C:

R₃₅₀ = R₀(1 + αΔT)

where,

R₃₅₀ = Resistance at 350°C = ?

α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹

ΔT = Difference in Temperature = 350°C - 20°C = 330°C

Therefore,

R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]

R₃₅₀ = 1.44 Ω

Now, for power at 350°C:

P₃₅₀ = VI₃₅₀

where,

P₃₅₀ = Power dissipation at 350°C = ?

V = constant potential difference = 0.64 volts

I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)

Therefore,

P₃₅₀ = V²/R₃₅₉

P₃₅₀ = (0.64 volts)²/(1.44 Ω)

P₃₅₀ = 0.28 watt

The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?

Answer:

29.96cm

Explanation:

Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.

Now using the lens equation as follows;

[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex]   -------------(i)

Where;

f = focal length of the lens

v = image distance as seen by the lens

u = object distance from the lens

From the question;

v = -151cm        [-ve since the image formed is virtual]

u = 25cm

Rewrite equation (i) to have;

[tex]f = \frac{uv}{u+v}[/tex]

Substitute the values of v and u into the equation;

[tex]f = \frac{25*(-151)}{25-151}[/tex]

[tex]f = \frac{-3775}{-126}[/tex]

f = 29.96cm

The focal length should be 29.96cm

Answer:

a) P = 10.27 kW

b) Pmax = 10.65 kW

c) E = 5.47 MJ

Explanation:

Mass of the loaded car, m = 950 kg

Angle of inclination of the shaft, θ = 28°

Acceleration due to gravity, g = 9.8 m/s²

The speed of the car, v = 2.35 m/s

Change in time, t = 14.0 s

a) The power that must be provided by the winch motor when the car is moving at constant speed.

P = Fv

The force exerted by the motor, F = mg sinθ

P = mgv sinθ

P = 950 * 9.8 *2.35* sin28°

P = 10,271.3 W

P = 10.27 kW

b) Maximum power that the motor must provide:

[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]

c) Total energy transferred:

Length of the track, d = 1250 m

[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]

Answer:

[tex]F=1.8\times 10^{-3}\ N[/tex]

Explanation:

We have,

Length of wires is 11 m

Separation between wires is 0.033 m

Current in both the wires is 5.2 A

It is required to find the magnitude of force between two wires. The force between wires is given by :

[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]

So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

Answer: 9.1 × 10^-26 Joule

Explanation:

Since the collision is elastic. The kinetic energy will be conserved. That is, the sum of kinetic energy before collision will be the same as the sum of the

energy after collision.

Mass of an electron = 9.1 × 10^-31 kg

Given that the velocity of the moving electron = 400 m/s and the stationary electron now has a velocity = 200 m/s. 

K.E = 1/2mv^2

Add the two kinetic energies

1/2mV1^2 + 1/2mV2^2

1/2m( V1^2 + V2^2 )

Since they both have common mass

Substitute m and the two velocities

1/2 × 9.1×10^-31( 400^2 + 200^2)

4.55×10^-31 ( 160000 + 40000 )

4.55×10^-31 × 200000

K.E = 9.1 × 10^-26 Joule

Therefore, the initial kinetic energy of the moving electron is 9.1×10^-26 J

Answer:

1.26 secs.

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Extention (e) = 0.2 m

Mass (m) = 4 Kg

Period (T) =.?

Next, we shall determine the spring constant, K for spring.

The spring constant, K can be obtained as follow:

Force (F) = 20 N

Extention (e) = 0.2 m

Spring constant (K) =..?

F = Ke

20 = K x 0.2

Divide both side by 0.2

K = 20/0.2

K = 100 N/m

Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:

Mass (m) = 4 Kg

Spring constant (K) = 100 N/m

Period (T) =..?

T = 2π√(m/K)

T = 2π√(4/100)

T = 2π x √(0.04)

T = 2π x 0.2

T = 1.26 secs.

Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.

Answer:

The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.

Explanation:

Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.

Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.

When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.

Answer:

F₁ = 100 N

Explanation:

The pressure must be equally transmitted from the piston to the narrow arm. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₁/F₂ = A₁/A₂

where,

F₁ = Force Required to be applied to piston = ?

F₂ = Force to push cork at narrow arm = 16 N

A₁ = Area of wider arm = πd₁²/4

A₂ = Area of narrow arm = πd₂²/4

Therefore,

F₁/16 N = (πd₁²/4)/(πd₂²/4)

F₁ = (16 N)(d₁²/d₂²)

but, it is given that the diameter of wider arm is 2.5 times the diameter of the narrow arm.

d₁ = 2.5 d₂

Therefore,

F₁ = (16 N)[(2.5 d₂)²/d₂²]

F₁ = (16 N)(6.25)

F₁ = 100 N

Answer:

MT disc   I = 2,752 10-3 kg m²

MB disc   I = 2,726 10⁻³ kg m²

Explanation:

The moment of inertia given by the expression

        I = ∫ r² dm

for bodies with high symmetry it is tabulated

  for a hollow disk it is

        I = ½ M (R₁² + R₂²)

let's apply this equation to our case

disc MT = 1,357 kg

         I = ½ 1,357 (0.0079² + 0.0632²)

         I = 2,752 10-3 kg m²

disk MB = 1,344 kg

         I = ½ 1,344 (0.0079² + 0.0632²)

         I = 2,726 10⁻³ kg m²

Answer:

C. explicit and implicit

Explanation:

E20

Answer:

e. any one of the above

Explanation:

Answer:

T = 7.61*10^6 s

Explanation:

In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:

[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex]         (1)

T: period of Mercury

r: distance between Mercury and Sun

Ms: mass of the sun = 1.98*10^30 kg

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

You replace the values of all parameters in the equation (1):

[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]

The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days

Answer:

Higher than the coefficient of kinetic friction.

Explanation:

Hope it helps u..   :)

Answer:

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Explanation:

Magnetic flux is the scalar product of the magnetic field over the area

               Ф = ∫ B. dA

where B is the magnetic field and A is the area

Let's look at stationary, for which factors affect flow

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Answer:

t = 0.22 s

Explanation:

The rate of curvature can be defined as the ratio of the radius of curvature to the time taken to bend the wire to that radius of curvature. Therefore,

v = r/t

where,

v = rate of curvature

r = radius of wire

t = time taken

Here, in our case:

v = 9 units/s

r = 2 units

t = ?

Therefore,

9 units/s = 2 units/t

t = (2 units)/(9 units/s)

t = 0.22 s

Therefore, it takes 0.22 second to bend a straight wire into a circle of radius 2 units