7. What was one advantage of the Stanford-Binet scale intelligence tests over previous ones?
The Stanford-Binet scale tests were more accurate.
The Stanford-Binet scale tests were given with paper and pencil versus conducting one-on-one interviews
The Stanford-Binet scale tests accounted for differences in social class.
The Stanford-Binet scale tests accounted for differences in test takers from different social classes.

Answers

Answer 1

Answer:

One advantage of the Stanford-Binet scale intelligence tests over previous ones is that they were given with paper and pencil, rather than conducting one-on-one interviews. This allowed for a more standardized and objective administration of the test, which helped to improve the reliability and validity of the test scores. Additionally, the Stanford-Binet scale tests were designed to account for differences in test takers from different social classes, which previous intelligence tests did not do. This made the Stanford-Binet scale tests more fair and unbiased, and allowed for more accurate comparisons between test takers.


Related Questions

Select the correct answer.
The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's motion?

A. It's not moving.

B.It's moving at a constant speed.

C.It's moving at a constant velocity

D.It's speeding up.

Answers

Answer:

It isn't moving

Explanation:

In hiking, what fitness component is required of you

Answers

It’s strength, endurance and flexibility. Hope this helps

how does the uneaven heating of earths surface affects earths weather patterns

Answers

Answer: it causes some parts of the earth to get more radiation than others.

Explanation: earth rotates around the sun on a tilted axis so the Rays of the sun cause earth to have more radiation than it needs.

Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons

Answers

Answer:

The electric potential energy of the system is 7.87x10⁻²⁰ J.

Explanation:

The electric potential energy is given by:

\(E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}\)

Where:

q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C

r is the distance = 3x10⁻⁹ m

K: is the electrostatic constant = 9x10⁹ Nm²/C²

\( E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J \)

Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.

I hope it helps you!

The electric potential energy of the system should be 7.87x10⁻²⁰ J.

Calculation of the electric potential energy:

SInce We know that

fdr = kq1q2/r

Here

q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C

r should be the distance = 3x10⁻⁹ m

K should be the electrostatic constant = 9x10⁹ Nm²/C²

Now electric potential energy should be

= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) /  3x10⁻⁹ m

=  7.87x10⁻²⁰ J.

hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.

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The first P-wave of an earthquake travels 5600 kilometers from the epicenter and arrives at a seismic station at 10:05 a.m. At what time did this earthquake occur?

Ahhhhhh I have a Regent's test in 2 hours and I don't know how to solve this type of question! Any help would be appreciated.

Anyone know what the steps to do this are? I dont even need an answer, just how to get to it. Thank you!

Answers

The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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What is the entropy change when 2 g of water changes from 0 °C to steam at 234 °C if (The thermal capacity is pressure) Cp = 1 J / kg.K and a latent heat of vaporization is 54 J/Kg?​

Answers

Answer:

\( \boxed{ \sf \: \Delta S = 0.219J /K}\)

Explanation:

Given:

mass of substance (water) = 2 g

initial temperature= 0°C = 273.15 K

final temperature= 234°C= 507.15

thermal capacity = 1 J/kg.M

latent heat of vaporization = 54 J/Kg

To find:

Entropy change ∆S = ?

Solution:

Entropy is measure of randomness of a system.

The change in entropy can be calculated in many ways depending upon the nature of substance & process like solid or liquid in isothermal process and when the process is not isothermal, gases are real or ideal etc.

If the process is isothermal (Temperature doesn't change) then the general equation to find change in entropy is,

\(\sf \Delta S = \frac{ q_{rev}}{T}= state \: function\)

If the temperature of reaction is changing then change in entropy can be calculated by

\( \sf \: \Delta S = nC_p ln \frac{T_f}{T_i} \)

Where n is number of moles of substance,

Cp is thermal capacity,

& T f and Ti are final and initial temperature respectively.

In the above reaction the water is being heated from 0°C (liquid) to 234°C (steam), the number of processes happening is

Entropy change (∆S1) for physical process where, water being heated from 0°C to 100°CEntropy change (∆S2) for phase transformation of isothermal process at 100°C where liquid H2O is turning into vapoursEntropy change (∆S3)for physical process, where vapours heated upto 234°C

Calculating the number of moles of substance,

Given mass of water= 2g

Molecular mass of water = 18 g

\( \sf n = \frac{given \: mass \:}{molar \: mass} = \frac{2}{18} = 0.11moles\)

Change in entropy for physical process 1,

The initial temperature of water is 0°C(273.15 K), and the water turns into vapour at 100°C (373.15 K), this is a purely heating of liquid at constant pressure, hence the formula that would be used is,

\(\sf \: \Delta S_1 = nC_p ln \frac{T_f}{T_i} \\ \sf \: \Delta S_1 = 0.11 \times 1 \times ln \frac{373.15}{273.15} \\ \sf \: \Delta S_1 = 0.11 \times ln 1.366 \\ \sf \: \Delta S_1 = 0.11 \times 0.311 \\ \sf \: \Delta S_1 = 0.0343 J/K\)

At boiling point of water the change in entropy for phase transformation of liquid water to gaseous vapours at 100°C can be calculated,

Now the latent heat of vaporization (∆H) is given,

(Latent heat is the heat which is required to change the phase of any substance)

Substituting the data,

\(\sf \: \Delta S_2 = \frac{\Delta H}{T} \\ \sf \: \Delta S_2 = \frac{54}{373.15} \\ \sf \: \Delta S_2 = 0.145 J/K\)

Now the third and final physical process of heating vapours from 100°C to 234°C, at constant pressure,

\( \sf \: \Delta S_3 = nC_p ln \frac{T_f}{T_i} \\ \sf \: \Delta S_3 = 0.11 \times 1 \times ln \frac{507.15}{373.15} \\ \sf \: \Delta S_3 = 0.11 \times ln 1.359 \\ \sf \: \Delta S_3 = 0.11 \times 0.306 \\ \sf \: \Delta S_3 = 0.0337J/K\)

The total entropy change ∆S during the process is,

\( \sf \: \Delta S = \Delta S_1+ \Delta S_2 + \Delta S_3 \\ \sf \: \Delta S = 0.0343+ 0.145 + 0.0337 \\ \sf \: \Delta S = 0.219J/K\)

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A cylinder contains 3.5 L of oxygen at 350 K and 2.7 atm . The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 620 K and the volume of the cylinder to increase to 9.1 L.What is the gas pressure? P= _____atm

Answers

Answer:

The pressure is \(P_2 = 1.84 \ a.t.m\)

Explanation:

From the question we are told that

   The first  volume of  is  \(v_1 = 3.5 \ L\)

   The first  pressure is  \(P_1 = 2.7 \ a.t.m\)

   The first  temperature is  \(T_1 = 350 \ K\)

    The  new temperature is  \(T_2 = 620 \ K\)

     The  new volume is  \(V_2 = 9.1 \ a.t.m\)

Generally according to the combined gas law we have that

      \(\frac{P_1 V_1 }{T_1 } = \frac{P_2 V_2 }{T_2 }\)

=>  \(P_2 = \frac{P_1 * V_1 * T_2 }{T_1 * V_2 }\)

=>    \(P_2 = \frac{ 2.7 * 3.5 * 620 }{ 350 * 9.1 }\)

=>  \(P_2 = 1.84 \ a.t.m\)

1.2miles=__________km

Answers

Answer:

1.931 kilometres is the answer of 1.2 miles

Answer and Explanation:

1 mile = 1.609 km

Set up a fraction to cancel the miles to get the kilometers.

\(\frac{1.2mi}{?km} *\frac{1.609}{1mi} = 1.9308km\) <- This is the answer.

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The rock has potential energy due to the gravitational potential with the Earth. As the rock is falling, this gravitational energy becomes less. What happens to that energy? Question 11 options: The energy disappears. The energy is transformed into kinetic energy. The energy is absorbed by the ground and the air The energy is transformed into chemical potential energy.

Answers

Answer:

Explanation:

while it is still falling it is turned into kinetic energy. While on the floor it turns onto potential energy. Remember energy cannot be created or the destroyed so the other ones are automatically out.

it is turned in to kinetic energy
kinetic energy is when something is moving

What is the difference between special relativity and general relativity? Briefly describe each theory and cite one piece of evidence supporting each theory.

Answers

Special relativity and general relativity are both theories proposed by Albert Einstein. Special relativity deals with the laws of physics in the absence of gravity, while general relativity extends special relativity to include gravity and explains the curvature of spacetime caused by mass and energy.

Special relativity, proposed in 1905, deals with the laws of physics in the absence of gravitational fields. It introduces the concepts of time dilation and length contraction, stating that the laws of physics are the same for all observers moving at constant speeds relative to each other.

One piece of evidence supporting special relativity is the famous Michelson-Morley experiment, which failed to detect the existence of the hypothetical luminiferous aether.

On the other hand, general relativity, formulated in 1915, is an extension of special relativity that incorporates gravity. It postulates that gravity arises from the curvature of spacetime caused by mass and energy. General relativity explains the motion of celestial bodies, the bending of light in the presence of massive objects, and phenomena like black holes.

One piece of evidence supporting general relativity is the observed gravitational redshift, where light emitted from a source in a strong gravitational field is shifted to longer wavelengths.

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The half-life of Silver-105 is 3.57 x 106 seconds. A sample contains 5.78 x 1017 nuclei. What is the decay constant for this decay?

Answers

Answer:

The decay constant, or "lambda" (λ), is the rate at which a radioactive isotope decays. It is usually measured in units of inverse time, such as seconds. In this case, the decay constant can be calculated as follows:

16:42

λ = (ln(2)/3.57 x 106) x (5.78 x 1017) = 0.

Explanation:

A body with mass m slides down a frictionless ramp inclined at 600, with an initial speed v1 = 3 m/s,
starting at height h, and continues its motion upward on a 300 inclined ramp with a friction. Than this
body hits a spring which is fixed to a wall and compresses it 10 cm. The spring is initially at a distance R
on horizontal plane, and at height h. Find:
a) The speed v2 at the point 0, where the body starts to climb up to 300 inclined ramp, in terms of h.
b) The speed v3 where it hits the spring.
c) The spring constant k.
m=0.1 kg, µk=0.1, R=2.5 m., v1=3 m/s, Cos(600) = 0.5, Sin(600)=0.87, x=10 cm. and g= 10 m/s2 .

Answers

Going down the first ramp:

• net force parallel to the ramp:

F = W sin(60°) = m a

(W for weight)

• net force perpendicular to the ramp:

F = N + W cos(60°) = 0

(N for normal force)

The body has mass 0.1 kg, and with g = 10 m/s², its weight is W = 1 N. So in the first equation, we get

(1 N) sin(60°) = (0.1 kg) a₁   →   a₁ ≈ 8.7 m/s²

Let d₁ be the distance the body moves down the ramp, i.e. the distance along the ramp between the starting point and the point O. Then

sin(60°) = h / d₁   →   d₁ = 2h/√(3) ≈ 1.15h

Given an initial speed v₁ = 3 m/s, we find the speed v₂ at point O to be

v₂² - (3 m/s)² = 2 (8.7 m/s²) d

v₂ ≈ √(9 m²/s² + (17.4 m/s²) (1.15h))

v₂ ≈ √(9 m²/s² + (20 m/s²) h)

Going up the second ramp:

• net parallel force:

F = -Fr - W sin(30°) = m a

(Fr for friction)

• net perpendicular force:

F = N - W cos(30°) = 0

sin(30°) = cos(60°), and cos(30°) = sin(60°), so the second equation gives N = 0.87 N. Then with µ = 0.1, we have Fr = µ N = 0.087 N. The first equation then gives

-0.087 N - 0.5 N = (0.1 kg) a₂   →   a₂ ≈ -5.9 m/s²

We now have

tan(30°) = h/R   →   h = (2.5 m)/√3 ≈ 1.4 m

(which, by the way, tells us that v₂ ≈ 6.2 m/s)

Then the distance traveled up the ramp is

d₂ = √(h² + R ²) ≈ 2.9 m

Use this to solve for the speed at the top of the ramp:

v₃² - v₂² = 2 (-5.9 m/s²) d

v₃ = √((6.2 m/s)² - (11.8 m/s²) (2.9 m)) ≈ 2.0 m/s

At the top of the second ramp:

• net parallel force:

F = -Fsp - W sin(30°) = m a

(Fsp for spring)

• net perpendicular force:

F = N - W cos(30°) = 0

By Hooke's law, Fsp = kx, so in the first equation we get

-k (0.10 m) - (1 N) cos(60°) = (0.1 kg) (-5.9 m/s²)

→   k ≈ 0.87 N/m

A body with mass m slides down a frictionless ramp inclined at 600, with an initial speed v1 = 3 m/s,

Select ALL the
correct answers.
Which two examples describe ways that corporations can give large donations to presidential candidates?
A corporation leader makes
direct payment to the candidate.
A corporation creates another company to accept candidate contributions
A corporation collects moneys from its employees to contribute to
A corporation contributes to a Super PAC that
a PAC
accepts contributions
for a candidate,
Reset
Next

Answers

Answer: Hope this helps ;) don't forget to rate this answer !

Explanation:

There are two correct answers:

A) A corporation leader makes direct payment to the candidate.

D) A corporation contributes to a Super PAC that a PAC accepts contributions for a candidate.

Option A describes a scenario where a corporation directly donates money to a presidential candidate, which is allowed as long as it is done within the limits set by campaign finance laws.

Option D describes a scenario where a corporation donates money to a Super PAC, which is a type of political action committee that can accept unlimited donations from individuals, corporations, and other organizations. The Super PAC can then use the money to support or oppose a particular candidate, but it is not allowed to coordinate directly with the candidate or the candidate's campaign.

I hope this helps! Let me know if you have any other questions.

A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.

Answers

Answer:

Explanation:

Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .

A )

Intensity of sound at 1 m distance = 60 /4 π d²

d = 1 m

Intensity of sound at 1 m distance = 60 /(4 π 1²)

= 4.78 W m⁻² s⁻¹

B )

Intensity of sound at 1.5 m distance = 60 /4 π d²

d = 1.5  m

Intensity of sound at 1 m distance = 60 /(4 π 1.5²)

= 2.12 W m⁻² s⁻¹

C )

Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²

d = 1.5  m

= 4 x 60 /(4 π 1.5²)

= 8.48 W m⁻² s⁻¹

D )

Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .

.06 /4 π d² = 10⁻¹²

d² = .06 /4 π 10⁻¹²

d = 6.9 x 10⁴ m .

what is light emitting diode

Answers

Answer:

A light emitting diode (a semiconductor diode which glows when voltage is applied.

Acceleration vs. Mass
16
The graph shows the acceleration of the cart as the
mass of the cart varies.
As the mass of the cart increases, the acceleration of
the cart increases
The type of relationship that exists between the mass of
the cart and the acceleration of the cart is
relationship.

Answers

Answer:

decreases, inverse

Explanation:

if you are on edge, i'm helping you, go run

To solve this we must be knowing each and every concept related to direct relationship and inverse relationships. Therefore, there is direct relationship that exists between the mass of the cart and the acceleration of the cart.

What is direct relationship?

In a direct connection, one variable's growth is directly proportionate to another's. Using the ball from the previous section as an example, a ball will bounce higher the higher it is dropped.

The way inverse relationships function varies. The quantity of y drops when x is raised. For instance, your travel time will be reduced if you go more swiftly to your destination. There is direct relationship that exists between the mass of the cart and the acceleration of the cart.

Therefore, there is direct relationship that exists between the mass of the cart and the acceleration of the cart.

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How many moles of aluminum atoms have a mass equal to the mass of a small airplane?.

Answers

The mass of a small airplane can vary widely, so we will assume a mass of 2,000 kg for the purposes of this calculation. To determine the number of moles of aluminum atoms that have a mass equal to this, we need to know the molar mass of aluminum, which is approximately 27 g/mol.

First, we need to convert the mass of the airplane to grams:

2,000 kg = 2,000,000 g

Next, we need to calculate the number of moles of aluminum atoms that have a mass equal to 2,000,000 g:

moles of aluminum = mass of aluminum / molar mass of aluminum

moles of aluminum = 2,000,000 g / 27 g/mol

moles of aluminum ≈ 74,074.07 mol

Therefore, approximately 74,074 moles of aluminum atoms have a mass equal to the mass of a small airplane with a mass of 2,000 kg.

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In (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.00 m to the right of its center. The beam weighs 155 N and makes an angle of 30.0∘ with the vertical. At the right-hand end of the beam a 100.0 N weight is hung; an unknown weight w hangs at the left end.If the system is in equilibrium, what is w ? You can ignore the thickness of the beam.

In (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.00 m to the right of its center.

Answers

Given::

Length = 6.00 m

Weight of beam = 155 N

Angle = 30.0 degrees

Weight added at the right-end = 100 N

If the system is in equilibrium, let's find the weight, w, at the left-end.

Since the system is in equilibrium, the net torque of the system will be zero.

Now, we have the equation:

\((100)(2.00sin30)-w(4.00sin30)-(155)(1.0sin30)=0\)

Where w is the weight at the left end.

Let's solve for w.

We have:

\(\begin{gathered} (100)(1)-w(2)-(155)(0.5)=0 \\ \\ 100-2w-77.5=0 \\ \\ 100-77.5-2w=0 \end{gathered}\)

Solving further:

\(\begin{gathered} 22.5-2w=0 \\ \\ 2w=22.5 \end{gathered}\)

Divide both sides by 2:

\(\begin{gathered} \frac{2w}{2}=\frac{22.5}{2} \\ \\ w=11.25\text{ N} \end{gathered}\)

Therefore, the weight hung at the left end is 11.25 N

ANSWER:

11.25 N

A cart moving across a level surface accelerates
uniformly at 1.0 meter per second² for 2.0
seconds. What additional information is
required to determine the distance traveled by
the cart during this 2.0-second interval?

Answers

The initial velocity of the cart.

Newton, There can be a mass is four.080, So acceleration might be equal to 2.50 m in step with cent within the rectangular. Initial is that amount that relies upon total mass. The greater mass the more inertia. So Mass is 2000 kg and acceleration is 3 ms square. So this offers us an internet pressure identical to 6000 newtons or 6.0 and 210 to the power

In case you roll a ball, it initially will keep rolling except friction or something else stops it by means of pressure. you could also think about the way that your body maintains transferring ahead when you hit the brake on your bike. Translational Inertia = ma, in which m is the mass, and a is the acceleration of the object. Calculate the rotational inertia or the instant of inertia velocity by way of multiplying the mass of the object with a square of the gap between the item and the axis, the radius of rotation.

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(Ic) Asmall ball carrying a charge of -3.0x10-12C experiences an east ward force of 8.0×10 N due to its charge when it is suspended at a certain point in space. What is the magnitude and direction of Eat that point? ​

Answers

The magnitude of the electric field is \(2.67*10^6 N/C\), and its direction is westward.

To determine the magnitude and direction of the electric field at the point where the small ball is suspended, we can use the formula for the electric field:

Electric field (E) = Force (F) / Charge (q)

Given:

Charge of the ball (q) = \(-3.0*10^{(-12)} C\)

Force experienced by the ball (F) = \(8.0*10^{(-6)}\) N (eastward)

Plugging in the values into the formula:

E = F / q

E = \((8.0*10^{(-6) }N) / (-3.0*10^{(-12)} C)\)

Let's calculate the electric field:

E = -\(2.67*10^6 N/C\)

The magnitude of the electric field is\(2.67*10^6 N/C\). The negative sign indicates that the electric field is pointing in the opposite direction to the force experienced by the ball, which in this case is westward.

Therefore, the magnitude of the electric field is \(2.67*10^6 N/C\), and its direction is westward.

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why aeroplanes cannot travel in space

Answers

Answer:

- They need oxygen to burn fuel

- Aerodynamics

- Extreme temperatures

- Radiation

- Pressure issues

Explanation:

A airplane is a heavier-than-air aircraft kept aloft by the upward thrust exerted by the passing air on its fixed wings and driven by propellers, jet propulsion, etc.

Aeroplanes cannot travel in space for several reasons:

They need oxygen to burn fuel - Aeroplane engines rely on the oxygen in the atmosphere to burn fuel and generate thrust. In space, there is no atmosphere so there is no oxygen for the engines to work.

Aerodynamics - Aeroplane wings generate lift by interacting with the air. In space, there is no air so wings would be unable to generate any lift. Aeroplanes rely on aerodynamics to fly which does not work in space.

Extreme temperatures - In space, temperatures can range from -150 degrees Celsius to 150 degrees Celsius. Aeroplanes are designed to operate within a much narrower temperature range. The extreme cold and heat of space could damage aeroplane components.

Radiation - In space, there are high levels of radiation from the Sun and cosmic rays. Aeroplane bodies are not designed to shield against this type of radiation and it could damage electronics and affect aeroplane systems.

Pressure issues - Aeroplanes are designed to withstand air pressures at altitudes up to around 12 kilometers. In low-Earth orbit and beyond, the air pressure is essentially zero. This extreme change in pressure could cause structural damage to the aeroplane.

In summary, while aeroplanes are designed to fly through the Earth's atmosphere, they lack the key features needed to operate in the extreme environment of outer space like spaceships. Aeroplanes require things like oxygen, aerodynamics and being able to withstand changes in pressure - all of which do not exist or work the same way in space.

Explanation:

The wing is pushed up by the air under it. Large planes can only fly as high as about 7.5 miles. The air is too thin above that height. It would not hold the plane up.

a certain compact disc (CD) contains 783.216 megabytes of digital information 4

Answers

Answer:

"a certain compact disc (CD) contains 783.216 megabytes of digital information 4"

most of our brains activity is:

Answers

most of our brain activity is unconscious

Imagine that the electrostatic forces on the cart and track were suddenly removed.When the cart hits the track, what evidence would there be that energy was transferred?

Answers

The change in motion, visible effects of the collision, and temperature increase would all serve as evidence that energy was transferred during the collision between the cart and the track after the removal of electrostatic forces.

If the electrostatic forces on the cart and track were suddenly removed and the cart hits the track, there would be several pieces of evidence indicating that energy was transferred during the collision.Firstly, there would be a noticeable change in the motion of the cart. The cart would decelerate as it collides with the track, and its velocity would decrease due to the loss of energy. This change in motion demonstrates that energy has been transferred from the cart to the track.

Secondly, there may be visible effects of the collision, such as deformation or damage to the cart or the track. This deformation or damage occurs because the energy transferred during the collision is transformed into other forms, such as heat or sound energy. These visible effects provide evidence of energy transfer.Additionally, there could be an increase in the temperature of the cart and/or the track due to the conversion of kinetic energy into thermal energy upon impact. This temperature change would be another indication that energy has been transferred.

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06.** A block takes twice as long to slide down an inclined plane that makes an angle of 35 with the horizontal as it does to fall freely through the same vertical distance h. What is the coefficient of kinetic friction? [μ = 0.16] ​

Answers

The coefficient of kinetic friction between the block and the incline is 0.168.

Coefficient of kinetic friction

The coefficient of kinetic friction is determined from the principle of conservation of energy as follows;

distance traveled on the incline = L

L = h/sin35

L = 1.743 h

Since it takes twice the time travel down,

velocity to slide = 1.743 h/2t = 0.871 v

v is the velocity to drop from the vertical height

K.E of the block when it slides = ¹/₂mv₁²  = ¹/₂m(0.871v)² = 0.871²(K.E to fall) = 0.758K.E to fall straight down

Energy lost to friction

1 - 0.758 = 0.24

Conservation of energy

μmgcos(35) = 0.24(mgsin(35)

μcos(35) = 0.24(sin(35)

μ = 0.168

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Write any 3 difference between work energy power​

Answers

Answer:

Work is referred to as the displacement of an object when a force (push or pull) is applied to it while energy is referred to as the capacity to do the work. It exists in various forms like potential, kinetic, chemical, thermal, nuclear, electrical energy and so on. Power is the work done per unit of time.

Types of Energy:Mechanical energy.Mechanical wave energy.Chemical energy.Electric energy.Magnetic energy.Radiant energy.Nuclear energy.Ionization energy.

power and energy are the same.
Work is usualy doing a task that gets rid of said energy.

if the Periodic time of an oscillating object Triples then its frequency will?​

Answers

Answer:

it would decrease

Explanation:

f=1/T

The amplitude of a wave
determines the volume of a
sound.
True
O False

Answers

true the amplitude of a wave determines the volume of a sound

An object changes velocity from 100m/s
to 76m/s in 6 seconds. If it has a mass of
20kg, what is the net force?

Answers

Answer:

1-1=0

Explanation:

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Convert 25 meters per second (m/s) to kilometers per hour (km/hr).
Оа
9.0 x 10^-2 km/hr

9.0 x 10^-1 km/hr
Ос
9.0 x 10^1 km/hr
Od
9.0 x 10^2 km/hr

Answers

Answer:

9 x 10^1  km/hr  

Explanation:

25 m/s   x  3600 sec / hr   x  1km / 1000 m = 90 km/hr

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