A surfactant with a Hydrophile-
Lipophile Balance (HLB) value of 18 is expected to function as a solubilizing agent
O False
O True​

Answer:

O Distant objects are blurry. describes farsightedness.

Explanation:

Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.

Answer:

The correct solution is "11.51 mA".

Explanation:

Given:

Time average power,

[tex]P_{avg}=0.05 \ W/m^2[/tex]

n = 377

As we now,

⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]

or,

⇒ [tex]E_0^2=0.05\times 377[/tex]

⇒       [tex]=4.341 \ V[/tex]

hence,

⇒ [tex]H_0=\frac{E_0}{n}[/tex]

By putting the values, we get

         [tex]=\frac{4.341}{377}[/tex]

         [tex]=11.51 \ mA[/tex]

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

Answer: C.

Explanation:  plato users

Answer:

Total cost = 56.16 cents

Explanation:

Given the following data;

Power = 45 Watts

Time = 4 hours

Number of days = 30 days

Cost = 10.4 cents

To find how much does it cost her to watch TV for one month;

First of all, we would determine the energy consumption of the TV;

Energy = power * time

Energy = 45 * 4

Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).

Energy consumption = 0.18 Kwh

Next, we find the total cost;

Total cost = energy * number of days * cost

Total cost = 0.18 * 30 * 10.4

Total cost = 56.16 cents

Answer:

4592.57 lb

Explanation:

The missing diagram for this question is attached in the image below.

Given that:

the weight of the car = 2890 lb

At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr

At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr

To ft/s:

[tex](V_A)[/tex]  = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

[tex](V_A)[/tex]  = 85.07 ft/s

[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

[tex](V_C)[/tex] = 26.4 ft/s

Between A to C, the total distance is;

[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]

Now, we need to determine the deceleration of the car using the formula:

[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]

[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]

[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]

[tex]696.96-7236.9049 = 2 a (654.154)[/tex]

[tex]-6539.9449 = 2 a (654.154)[/tex]

[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]

a = -4.99 ft/s²

The velocity of the car as it passes via B

[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]

[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]

[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]

[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]

[tex]v_B =\sqrt{ 10540.2849}[/tex]

[tex]v_B =102.67 \ ft/s[/tex]

Along B, the car's acceleration is:

[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]

[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]

[tex]a_B = 51.17 \ ft/s^2[/tex]

Finally, the total horizontal force F exerted = m[tex]a_B[/tex]

[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]

= 4592.57 lb

Answer:

Road A- dry

Road B- mud

Road C- wet

Explanation:

Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.

The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.

The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.

a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.

b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.

Given the following data:

a. To determine the force exerted by the ground on each of the front wheels:

First of all, we would take moment about the rear wheels.

[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]

Force, F = 9836.32 Newton

For each front wheel:

[tex]Force = \frac{9836.32}{2}[/tex]

Force = 4918.16 Newton.

b. To determine the force exerted by the ground on each of the back wheels:

We would determine the sum of the vertical forces acting on the wheels.

[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]

B = 5255.68 Newton.

For each back wheel:

[tex]Force = \frac{5255.68}{2}[/tex]

Force = 2627.84 Newton.

Read more: https://brainly.com/question/22210180

The question is incomplete, the complete question is;

In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?

A) The atom moves to a state of lower energy

B) The atom is ionized

C) One of the electrons leaves the atom

D) The atom can be excited to a higher energy state

Answer:

The atom can be excited to a higher energy state

Explanation:

According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.

If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.

Rather, the atom is excited from ground state to a higher energy level.

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

substituting the values of the variables into the equation, we have

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

Answer:

46

Explanation:

Answer:

a) B = 3.11 km.  θ= 54.7º E of S

b) B = 3.11 km  θ= 54.7º W of S

Explanation:

a)

        [tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]

       [tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]

        [tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]

b)

      [tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]

Answer:

Tenemos dos problemas a resolver acá:

Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.

Acá podemos usar la conservación de la energía.

E = U + K

U = energía potencial = m*g*H

m = masa

g = aceleración gravitatoria = 9.8m/s^2

H = altura

K = energía cinética = (m/2)*V^2

donde V es la velocidad.

Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:

Ei = U = m*(9.8m/s^2)*2.5m

Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:

Ef = (m/2)*V^2

Y como la energía se conserva, la energía final es igual a la inicial, entonces:

m*(9.8m/s^2)*2.5m = (m/2)*V^2

Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.

V = √(2*(9.8m/s^2)*2.5m) = 7m/s

Ahora respondamos la segunda parte.

Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:

A(t) = -9.8m/s^2

Para obtener su velocidad integramos:

V(t) = (-9.8m/s^2)*t + V0

donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s

V0 = (3/4)*7m/s = (21/4) m/s

Así, la ecuación de la velocidad es:

V(t) = (-9.8m/s^2)*t + (21/4) m/s

Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:

V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s

         t =  (21/4) m/s/9.8m/s^2 = 0.54 s

Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.

Para ello integramos de vuelta:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0

donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:

P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t  

La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s

P(0.54s) =  (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m

La altura máxima es 1.81 metros.

Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.

Answer: When it comes to the movement of air, friction is greater near the ground surface.

Explanation:

A resistance in motion observed by an object while on another object is called friction.

For example, a vehicle moving on road will have friction between its tires and the road.

Friction is more near the ground surface rather than away from the ground surface.

Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.

Answer:

wait

Explanation:

Answer:

a) 10.3 m/s

b) 566 N

Explanation:

[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]

[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]

The athelete velocity will be 10.3 and constant force 566 N.

The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).

Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.

The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.

Therefore, The athelete velocity will be 10.3 and constant force 566 N.

To learn more about velocity, refer to the link:

https://brainly.com/question/18084516

#SPJ5

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]

Now, the distance covered by the player in this time will be:

[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]

s₁ = 0.022 m

Answer:

parallel circuit

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;

I. Series circuit

II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.

Hence, the type of circuit that the above diagram above represents is a parallel circuit.

Answer:

parallel circuit

Explanation:

I got it right on my exam

Answer:

amperes

Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.

Explanation:

hope it helps

Answer:

a. The disk

b. Because it has the smallest rotational inertia

Explanation:

a. Which object do you expect to reach the bottom of the inclined plan first?

I would expect the disk to reach the bottom first.

b. Why?

This is because the disk has the smallest rotational inertia.

The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.

Since the three objects are rolling from the same height, they have the same mechanical energy.

But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy.  The disk's smaller rotational kinetic energy will cause  to rotate less but translate more than the other objects and thus reach the bottom first.

The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.

Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.

There are three objects, hollow sphere, disk and ring.

       [tex]I=\dfrac{2}{3}mr^2[/tex]

        [tex]I=mr^2[/tex]

Here, (m) is the mass and (r) is the radius of the object.

These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.

As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.

Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.

Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.

Learn more about the force of inertia here;

https://brainly.com/question/10454047

Answer:

Ray A - incident ray

Ray B - reflected ray

Answer:

below

Explanation:

Answer:

D. Non- polar solvant

Explanation:

l think that's it

Answer:

I think the answer is D polar solvent

Answer:

144 meters

Explanation:

the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t

S (height cliff)=24m/s×6s=144

Answer:

8 N

Explanation:

Applying,

(F'+F) = ma............... Equation 1

Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.

From the question,

Given: F = 12 N, m = 4 kg, a = 5 m/s²

Substitute these values into equation 1

(F'+12) = 4×5

(F'+12) = 20

F' = 20-12

F' = 8 N.

Hence Ahmed's force is 8 N

Answer:

O The environment did work on an object

Explanation:

(a) Attached to the response as Figure 1.

(b) 35.0Ω

(c) Across 5.0Ω = 1.3V

   Across 10.0Ω = 2.6Ω

   Across 20.0Ω = 5.2Ω

(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.

(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e

R = R₁ + R₂ + R₃               -------------------(i)

Where;

R₁ = 5.0 Ω

R₂ = 10.0 Ω

R₃ = 20.0 Ω

Substitute these values into equation (i) as follows;

∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω

∴ R = 35.0 Ω

Therefore, the equivalent resistance is ∴ R = 35.0Ω

(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;

i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.

ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;

V = I R              ------------------(ii)

Where;

V = Voltage supplied to the circuit = 9.0V

I = Current through the circuit

R = Resistance of the equivalent resistor = 35.0Ω

Substitute these values into equation (ii)

9.0 = I x 35.0

I = [tex]\frac{9.0}{35.0}[/tex]

I = 0.26A

This is also the current flowing through each of the resistors separately.

iii. Calculate the voltage drop across

1. 5.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 5.0Ω resistor

I = current through the 5.0Ω resistor = 0.26A

R = resistance of the 5.0Ω resistor = 5.0Ω

=> V = 0.26 x 5.0

=> V = 1.3V

2. 10.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 10.0Ω resistor

I = current through the 10.0Ω resistor = 0.26A

R = resistance of the 10.0Ω resistor = 10.0Ω

=> V = 0.26 x 10.0

=> V = 2.6V

3. 20.0 Ω resistor

Applying Ohm's law from equation (ii)

V = I x R

Where;

V = voltage drop across the 20.0Ω resistor

I = current through the 20.0Ω resistor = 0.26A

R = resistance of the 20.0Ω resistor = 10.0Ω

=> V = 0.26 x 20.0

=> V = 5.2V

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

[tex]2gh = v_f^2 - v_i^2[/tex]

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]

(c)

Now we will consider the downward motion and use the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

Answer:

  a₃ = -1.08 m/s²,    K = 1.42 J

Explanation:

The particle is in a periodic motion, so the general expression is

           x = A cos (wt + Ф)

let's compare the terms with the expression they give us

           x = 2 cos (t - 2)

the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.

to find the acceleration we use its definition

          v = dx / dt

          a = dv / dt

          a = [tex]\frac{ d^2x}{dt^2}[/tex]

let's perform the derivative

          v = - A w sin (wt + Ф)

          a = - A w² cos wt + Ф)

substituting the values

          a = - 2 1² cos (t-2)

for t = 3 s

          a₃ = 2 cos (3-2)

remember angles are in radians

          a₃ = -1.08 m/s²

To calculate kinetic energy, let's find the velocity for t = 3 s

          v = - 2 sin (t-2)

          v = -2 sin (3-2)

          v = - 1.683 m / s

body mass is m = 1 kg

we calculate

          K = ½ m v²

          K = ½ 1 (-1.683) ²

          K = 1.42 J