PLEASE HELP ON THIS QUESTION ​

Answer:

D

Explanation:

Answer:

do it got a picture

on the edge

Explanation:

Answer:

Explanation:

I hope it's helpful for you

Randall has an unconscious assumption that attractive people are more competent.

The term assumption can be described as an unspoken premise that underlies the conclusion.

Here,

The capacity to accurately evaluate and draw conclusions about other individuals based on their overall physical appearance, verbal behaviour, and nonverbal attitudes is referred to as social perception.

Given that, for his restaurant, Randall is employing chefs. The first applicant is a dashing man with an average resume and career history. The second candidate is a far less appealing man with an average to slightly above average resume and work experience. Randall chooses to hire the first candidate.

This shows the social perception of Randall and his unconscious assumptions against unattractive people.

Hence,

Randall has an unconscious assumption that attractive people are more competent.

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Answer:

3) 311 m

Explanation:

Circumference = 2πR = π m/rev

99 rev(π m/rev) = 99π m or about 311 meters

The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]

Given the following data:

Radius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]

To find the centripetal acceleration of this ball:

The acceleration of an object along a circular track is referred to as centripetal acceleration.

Mathematically, the centripetal acceleration of an object is given by the formula:

[tex]A_c = \frac{V^2}{r}[/tex]

Where:

V is the velocity of an object.

Substituting the given parameters into the formula, we have;

[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]

Centripetal acceleration = 12 [tex]m/s^2[/tex]

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Answer:

200 feet

Explanation:

Answer:

The data suggest that the true mean HAZ depth is larger when the current setting is higher.

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dissimilar with respect to one or more particular characters

Answer:

WD = 960 J

Explanation:

WD = work done (J)

F = force (N)

s = displacement (m)

m = mass (kg) = 60

a = acceleration (m/s²) = 2

t = time (s) = 4

u = initial velocity (m/s) = 0

The formulas or equations that are relevant ate:

WD = F × s

F = m × a

s = u + at

We want to find WD, so we need to now the force and the displacement (or distance);

We calculate force, in Newtons, with the formula F = ma:

F = 60 × 2

F = 120 N

We also need displacement, which get with the formula s = u + at:

s = 0 + 2(4)

s = 8 m

Now we have F and s, we can calculate WD:

WD = 120 × 8

WD = 960 J

Methodology:

Starting with what you want to find, in this case WD, list the formula/s you could use;

Then, identify the information you need for the formula and whether or not you are given that information;

Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;

Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;

I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.

Total liquid: 73 + 25 = 98 ml

Percent juice = (25/98) x 100 = 25.5 %

We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."

Answers:

From the question we are told

The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s

A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube

Therefore, the mass flow rate of CCI_4 at point A

=  [tex]5.86*10^{-13} kg/s[/tex]

B) From Fick's law

[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]

Then,

[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]

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The moment of inertia of the wheel is 4.27 kg.m²

The kinematics equation explains the variables associated and related of motion.

From the information given, applying the kinematic equation of motion to determine the acceleration of the block, we have:

[tex]\mathbf{y = ut + \dfrac{1}{2}at^2}[/tex]

[tex]\mathbf{y = (0)t + \dfrac{1}{2}at^2}[/tex]

[tex]\mathbf{y = \dfrac{1}{2}at^2}[/tex]

Making acceleration (a) the subject, we have:

[tex]\mathbf{a = \dfrac{2y}{t^2}}[/tex]

where;

[tex]\mathbf{a = \dfrac{2\times 1.5 }{2.0^2}}[/tex]

a = 0.75 m/s²

The angular acceleration of the wheel can be estimated by the formula:

[tex]\mathbf{\alpha = \dfrac{a}{r}}[/tex]

[tex]\mathbf{\alpha = \dfrac{0.75 \ m/s^2}{0.40 \ m}}[/tex]

[tex]\mathbf{\alpha = 1.875 \ rad/s^2}[/tex]

Finally, the torque acting on the wheel is:

[tex]\mathbf{\tau = I \alpha}[/tex]

[tex]\mathbf{Tr = I \alpha}[/tex]

where;

∴

[tex]\mathbf{I =\dfrac{T\times r}{\alpha} }[/tex]

[tex]\mathbf{I =\dfrac{20 \ N\times 0.40 \ m}{1.875 \ rad/s^2} }[/tex]

I = 4.27 kg.m²

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After time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]

The given parameters;

The position function of the ball after time t, is calculated as follows;

[tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]

The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.

Thus, after time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]

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Answer:

c.

Explanation:

If the two plates that meet at a convergent plate boundary both are of oceanic crust, the older, denser plate will subduct beneath the less dense plate. The older plate subducts into a trench, resulting in earthquakes. Melting of mantle material creates volcanoes at the subduction zone.

When two oceanic plates converge, the denser plate will end up sinking below the less dense plate, leading to the formation of an oceanic subduction zone. Old, dense crust tends to be subducted back into the earth. An example of a subduction zone formed from a convergent boundary is the Chile-Peru trench….

Answer:

a divergent plate boundary

This question involves the concepts of the law of conservation of momentum, velocity, and mass.

The two quantities, the students should multiply before and after the collision are "A. mass and velocity".

According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.

To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.

This situation is illustrated in the attached picture.

So, according to the law of conservation of momentum:

Total Momentum Before Collision = Total Momentum After Collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.

NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!

1. Mass & velocity

2. The total momentum after the collision is the same as the total momentum before the collision.

3. 0.54 kg⋅m/s

4. The system has external forces, such as friction and air resistance, acting on it.

5. 3.0 m/s

Answer:

Explanation:

If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change

Answer:

Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.

Explanation:

Answer:

South-North

Explanation:

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]

The steepness of the slope is therefore;

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]

Where;

[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m

[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]

[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]

[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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The definition of volume modulus and the variation of pressure with depth allows to find the result for the variation of the volume of the coin is:

The pressure with the depth is given by the relation

         P = P₀ + ρ g h

Where P is the pressure, ρ is the density anf h depth.

The size of the bodies is determined by the distance of their atomic and molecular bonds, therefore the size of the bodies changes under external interations, in the case of hydrostatic pressure a constant called volumetric modulus is defined.

            [tex]B = - \frac{\Delta P}{\frac{\Delta V}{Vo} } \\\Delta V = - \frac{\Delta P }{B} \ V_o[/tex]

Where ΔP is the pressure change, V₀ and V are the volume change and the initial volume of the body, the negative sign is introduced so that the volumetric modulus is a positive quantity.

They indicate the diameter and thickness of the coin (d = 6.1 cm and e =0.20 cm) on the sea surface and the depth to which it is submerged

h = 770 m

Let's look for the volume of the coin.

          V₀ = π r² h = [tex]\pi \ \frac{d^2}{4} \ e[/tex]  

          V₀ = [tex]\pi \ \frac{0.061^2 }{4} \ 0.002[/tex]  

          V₀ = 5.84 10-6 m³

Let's find the pressure at the depth of y = 770 m,  the density of sea water is ρ = 1025 kg / m³, the pressure at the surface is the atmospheric pressure P₀ = 1 10⁵ Pa, the volumetric modulus of water is B = 0.21 10¹⁰ Pa.

          P = 1 10⁵ + 1025 9.8 770

          P = 1 10⁵ + 7,735 10⁶

          P = 7.84 10⁶ Pa

Let's calculate

          ΔV =[tex]- \frac{1 \ 10^5 - 7.84 \ 10^6 }{0.21 \ 10^{10}} \ 5.845 \ 10^{-6}[/tex]  

          ΔV = 2.15 10-8 m³

In conclusion using the definition of volume modulus and the variation of pressure with depth we can find the result for the variation of the volume of the coin is:

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Answer:

If the observer is stationary but the source moves toward the observer at a speed vs, the observer still intercepts more waves per second and the frequency goes up. This time it is the wavelength of the wave received by the observer that is effectively shifted by the motion, rather than the speed.

Answer:

20 m/s at -35°

Explanation:

Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.

Answer: Are these free point?

Explanation:

Answer:

Enrol in our 50 studyscore masterclass. Click here! Fuel cells are a type of primary cell in that they are not recharged, However, they are unique because they never run out, if the reactants are constantly supplied.

Answer:

Planck postulated that the energy of light is proportional to the frequency, and the constant that relates them is known as Planck's constant (h). His work led to Albert Einstein determining that light exists in discrete quanta of energy, or photons.

Explanation:

Answer:

Energy does not occur in continuous amounts but in discrete amounts described by:

E = N h ∨   where N is the number of quanta (energy units), ∨ the frequency of the energy, and h Planck's constant (6.63E-34 J-sec)