A student conducts an investigation to determine how the force of gravity affects different objects dropped from different heights. The student tests each object one time and announces that all objects experienced gravity the same way.

What is wrong with the student’s reasoning?


A.Gravity is unable to be tested with simple experiments, but only in a science lab.

B.The student failed to conduct repeated investigations with the same object to test for result reliability.

C.People are unable to test the effect of gravity on an object because the force is too strong, so the results were invalid.

D.The student failed to create a graph or chart to communicate the results instead of just announcing them.

The mass of a substance is not directly related to latent heat. Instead, latent heat is a parameter that describes the amount of energy required or released during a phase shift of a substance.

Latent heat can be thought of as hidden energy that is supplied or extracted to change the state of a substance without changing its temperature or pressure.

Latent heat is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process—typically a first-order phase transition.

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Answer:

The appropriate alternative is Option B (0.105 sec.).

Explanation:

The given values are:

Elastic modulus,

Y = 9.10 × 10¹⁰ N/m²

Mass,

m = 10.0 kg

Length of rod,

l = 2.00 m

Diameter,

d = 1.00 mm

Now,

⇒ \(Keq=\frac{AY}{l}= \frac{\pi D^2 Y}{4l}\)

On substituting the values, we get

⇒                   \(=\frac{\pi \times 10^{-6}\times 9.1\times 10^{10}}{4\times 2}\)

⇒                   \(=3.574\times 10^4\)

The time period will be:

⇒ \(T=2\pi \sqrt{\frac{m}{Keq} }\)

On substituting the above values, we get

⇒     \(=2\pi \sqrt{\frac{10}{3.574\times 10^4}}\)

⇒     \(=0.105 \ seconds\)

The time period resulting oscillations will be 0.1005 seconds.

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The given data in the problem is;

\(\rm \gamma\) is the elastic modulus=9.10 × 10¹⁰ N/m²

m is the mass= 10.0 kg

l is the length of brass rod= 2.00 m

d is the diameter of 1.00 mm

The value of the equivalent stiffness will be;

\(\rm K_{eq}= \frac{AY}{l}\\\\ \rm K_{eq}= = \frac{\pi d^2 Y}{4l} \\\\ \rm K_{eq}=\frac{3.14 \times 10^{-6}\times 9.1 \times 10^{10} }{4\times 2 } \\\\ \rm K_{eq}= 3.574 \times 10^4\)

The time period of the  oscillation is given by;

\(\rm T = 2 \pi \sqrt{\frac{m}{k_{eq}} } \\\\ \rm T = 2 \times 3.14 \sqrt{\frac{10}{3.574 \times 10^4}\)

\(\rm T = 0.105 \ sec\)

Hence the time period resulting oscillations will be 0.1005 seconds.

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Answer:

B. Mass would be the same but its weight would be less on the Moon.

Explanation:

The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.

Thus, the mass of a body is constant either on the Earth or on the Moon. But the weight would be less on the Moon because the gravitational force on the Moon is far less than that on the Earth. Therefore the weight would be less on the Moon.

The appropriate option is B.

The mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.

The prime focus to solve this problem is the mass and weight of an object. The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.

So, the mass of a body is constant either on the Earth or on the Moon. But the weight of an object will depend on the mass and the gravitational acceleration.

W = mg

Here, W is weight, m is mass and g is gravitational acceleration.

Weight would be less on the Moon because the gravitational force on the Moon is far less (due to lower value of g) than that on the Earth. Therefore the weight would be less on the Moon.

Thus, we can conclude that the mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.

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Answer:

Salt water is more dense than freshwater

Explanation:

The only and best explanation for this phenomenon for this occurrence is that salt water is more dense than freshwater.

If a 5kg mass starts from rest at xo = -1 and moves under the action of a variable force F(x) = √1-x² to point xf = 1. Then the total work done by the force is equal to π/2 + 1.

To calculate the total work done by the force in this scenario, we can use the formula for work:

Work = ∫F(x) dx

where F(x) is the force as a function of position and dx represents an infinitesimal displacement.

In this case, the force is given by F(x) = √(1 - x²), and we need to find the total work done as the object moves from xo = -1 to xf = 1.

Let's break down the calculation step by step:

Write the integral for work:

Work = ∫F(x) dx

Substitute the given force:

Work = ∫√(1 - x²) dx

Integrate with respect to x:

To integrate the square root of (1 - x²), we use the trigonometric substitution. Let's substitute x = sin(θ) and dx = cos(θ) dθ.

Work = ∫√(1 - sin²(θ)) cos(θ) dθ

Simplify the integrand:

Using the trigonometric identity sin²(θ) + cos²(θ) = 1, we can rewrite the integrand as cos²(θ).

Work = ∫cos²(θ) dθ

Apply the power-reducing formula:

The power-reducing formula states that cos²(θ) = (1 + cos(2θ)) / 2. We can use this formula to simplify the integrand further.

Work = ∫(1 + cos(2θ))/2 dθ

Integrate the terms separately:

Work = (1/2) ∫dθ + (1/2) ∫cos(2θ) dθ

The first integral, ∫dθ, is simply θ, and the second integral, ∫cos(2θ) dθ, can be calculated as sin(2θ)/2.

Work = (1/2) θ + (1/2) (sin(2θ)/2) + C

Evaluate the integral limits:

To find the total work done, we need to evaluate the integral at the upper and lower limits of integration.

At xf = 1, the angle θ is π/2, and at xo = -1, the angle θ is -π/2.

Work = (1/2) (π/2) + (1/2) (sin(2(π/2))/2) - [(1/2) (-π/2) + (1/2) (sin(2(-π/2))/2)]

Simplifying further:

Work = π/4 + (1/2) - (-π/4 + (1/2))

Work = π/4 + 1/2 + π/4 + 1/2

Work = π/2 + 1

Therefore, the total work done by the force is equal to π/2 + 1.

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The linear density of the string is 2.86 x 10^-5 kg/m.

What is linear density?

Linear density, also known as linear mass density, is a measure of the mass of a material per unit length. It is typically used to describe the properties of thin, elongated objects such as strings, wires, and fibers. The unit of measurement for linear density is typically kilograms per meter (kg/m) or grams per centimeter (g/cm). It is calculated by dividing the mass of the material by its length.

To find the linear density of the string, we need to use the relationship between the tension, linear density, and wave properties, which is given by the equation:

Tension = (linear density) * (wave speed)^2

where wave speed is the product of the frequency and wavelength of the wave.

Given the tension, period and wavelength, we can find the linear density using the following steps:

frequency = 1 / period = 1 / 3.82 x 10^-3 s = 261.9 Hz

wave speed = wavelength * frequency = 1.26 m * 261.9 Hz = 331.9 m/s

Tension = (linear density) * (wave speed)^2

944 N = (linear density) * (331.9 m/s)^2

linear density = Tension / (wave speed)^2 = 944 N / (331.9 m/s)^2 = 2.86 x 10^-5 kg/m

Therefore, the linear density of the string is 2.86 x 10^-5 kg/m.

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Bonjour,

AU is the astronomical unit referring to the distance between the Earth and the Sun 1 AU is equal to 1,50 × 10¹¹ m.

It therefore comes,

\(\Delta t = \frac{d}{v} = \frac{9,5 \times 1,5 \times 10^{11}}{3,0 \times 10^8} = 4 750 \Longleftrightarrow \Delta t \approx 80 \text{ m}\)

Answer:

19,200j

Explanation:

formular W=f×d

a) W=240N×20m

= 4,800j

b) W=200N×20m

=4000j

C) Td= 4800×4000\20×50

td= 19,200,000/1000

td= 19,200j

sorry if am wrong ☹️

(a) When m₂ is released, its acceleration will be approximately -3.27 m/s².

(b) The tension in the string is approximately -13.08 N.

To determine the acceleration of m₂ when it is released and the tension in the string, we need to consider the forces acting on the system.

(a) Acceleration of m₂:

Since the pulley is assumed to be frictionless, the tension in the string is the same on both sides of the pulley. We can consider the system consisting of m₁ and m₂ as one body. The net force acting on this system is the difference between the weight of m₁ and the weight of m₂:

Net force = m₁g - m₂g

Applying Newton's second law, F = ma, where F is the net force and a is the acceleration, we have:

m₁g - m₂g = (m₁ + m₂)a

Rearranging the equation to solve for the acceleration, we get:

a = (m₁g - m₂g) / (m₁ + m₂)

Substituting the given values, m₁ = 2.00 kg and m₂ = 4.00 kg, and the acceleration due to gravity, g = 9.8 m/s², we can calculate the acceleration:

a = ((2.00 kg)(9.8 m/s²) - (4.00 kg)(9.8 m/s²)) / (2.00 kg + 4.00 kg)

a = (19.6 N - 39.2 N) / 6.00 kg

a = -19.6 N / 6.00 kg

a = -3.27 m/s²

Therefore, when m₂ is released, its acceleration will be approximately -3.27 m/s². The negative sign indicates that the acceleration is in the opposite direction of the gravitational force.

(b) Tension in the string:

The tension in the string can be determined by considering the forces acting on m₂. The net force on m₂ is equal to its mass multiplied by its acceleration:

Net force = m₂a

Substituting the given values, m₂ = 4.00 kg and a = -3.27 m/s², we can calculate the tension:

Tension = (4.00 kg)(-3.27 m/s²)

Tension = -13.08 N

Therefore, the tension in the string is approximately -13.08 N. The negative sign indicates that the tension acts in the opposite direction of the weight of m₂.

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Answer:

Options B

Explanation:

The appropriate quantitative research method for understanding political views held by the young population of a specific area is written surveys (option b). Surveys allow for the collection of data from a large number of participants, and specific questions can be asked to gather data on political views.

Participant observation (option a) involves direct observation of individuals in a natural setting, which may not be practical for studying political views.

Secondary data analysis (option c) involves analyzing data that has already been collected, and may not be specific to the young population or the area of interest.

Laboratory experiments (option d) are typically used to study cause-and-effect relationships between variables, which may not be applicable to studying political views.

Therefore, the best option for understanding the political views held by the young population of a specific area is written surveys.

To understand the political views of the young population of a specific area, a researcher can use written surveys, participant observation, and secondary data analysis as quantitative research methods.

If a researcher is trying to understand the political views held by the young population of a specific area, they should use written surveys, participant observation, and secondary data analysis as quantitative research methods.

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Answer:

Option D, only on the portion of the Earth facing directly toward the Moon

Explanation:

Tides are caused by the gravitational pull of moon. The part of earth that faces the moon experiences the highest gravitational force and hence the high tides will occur in this regions only. The regions that do not faces the moon experiences low tides. It is the gravity of moon that attracts the ocean water towards itself.

Hence, Option D is correct

Answer:

a)The direction of the impulse that the glass imparts to the baseball Is opposite the direction of the balls motion

b)  \(I=0.6336Ns\)

Explanation:

From the Question we are told that:

Mass \(m=0.144kg\)

Initial Speed \(v_1=14.9m/s\)

Final speed  \(v_2=10.5\)

a)The direction of the impulse that the glass imparts to the baseball Is opposite the direction of the balls motion

b)

Generally the equation for impluse magnitude is mathematically given by

 \(I=m(v_1-v_2)\)

Therefore

 \(I=0.144(14.9-10.5)\)

 \(I=0.6336Ns\)

The well that will have most of the water will be well A.

The underground water supply is defined as a type of water that exists underground in saturated zones beneath the land surface.

From the two wells represented in the diagrams above, Well A has water supply from underground which is lacking in well B.

Therefore, well A will have most of the water more than B.

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Answer:

Explanation:

I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.

I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.

Potential energy at the start of a 2.0 m slide

PE = mgh = mg2sin30 = mg2(½) = mg J

0.8mg J gets converted to heat and 0.2mg converts to kinetic energy

0.2mg = ½mv²

v² = 0.4g

v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s

a) The value of t u = 140/t`b.

b) The y position of the cannonball at the time t is  55.5 mc.

c) The initial speed of the projectile is 52.4 m/s.

Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.

a. To find the value of t:Here, we have to use the formula of distance

i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`

Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is

(2H) = 2 × 55.5

= 111 m`2H/g

= 111/9.8`

= 11.32653 s

From equation (1),u×t = 140u = 140/t

Therefore, `u = 140/t`b.

b)To find the y position of the cannonball at the time t:

Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,

Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².

We have already calculated the time taken by the cannonball to hit the ground in the previous step.`

y = 140 - (1/2) × 9.8 × t²`

On substituting the value of t as `t = 11.32653`,

we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.

c)  To find the initial speed of the projectile:

To calculate the initial speed of the projectile, we need to use the formula of range of projectile

.i.e.,R = u²sin2θ/g

Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)

u² = R × g/sin2θ

   = 140 × 9.8/sin0°  

   = 2744m²/s²u  

   = \(\sqrt(2744m^2/s^2)\)

   = 52.4 m/s

Hence, the initial speed of the projectile is 52.4 m/s.

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Answer:

what are we supposed to find? can you make it clear?

The acceleration and distance:

a2 =  2,168 m/s²

In general, the following tactic constitutes the overall strategy:

(I) To begin, you will need to determine the child's velocity v at the point of impact.

(II) after that, determine the amount of deceleration based on the stopping distance.

These two equations, which originate from the explanation of what it means to "accelerate," are used in both calculations:

v = at (1) (1)

s = at²/2 (2) (2)

where

a stands for acceleration.

t is the amount of time needed to go from having no velocity to having a certain velocity, or vice versa.

The value s represents the total distance covered during the allotted time t.

Let's get ready to perform the algebraic operations that are listed below.

From (2) t = √(2s/a)                                            (3)

From (1) and (3) v = a√(2s/a) = √(2sa)               (4)

From (4) a = v²/2s                                               (5)

Let

s1 = 0.42 m be the distance of the fall,

s2 = 1.9 mm or 1.2 cm be the stopping distance,

t1 will be an unknown value representing the amount of time it takes to fall,

t2 will be an unknown value representing the amount of time it takes to decelerate, v will be an unknown value representing the velocity at impact, and a1 will be the gravitational acceleration (experienced during fall)

a2 is an unknown value that represents the deceleration felt during the impact.

(I)

To calculate v, use equation (4) substituting s = s1, a = a1

Then

v = √(2s1a1)                                                                                  (6)

(II)

To calculate a2 and t2, substituting s = s2, a = a2, and v from (6).

Then using (5) and (6)

a2 = v²/2s2 = 2s1a1/2s2 = (s1/s2)a1

Using (3)

t2 = √(2s2/a2) = √(2s2/((s1/s2)a1) = √(2s2²/(s1a1) = s2√(2/(s1a1))

Now substitute actual numbers:

The case of hardwood

a2 = (0.42/0.0019)×9.81 = 2,168 m/s²

t2 = 0.0019×√(2/(0.42×9.81) =  0.0013 s = 1.3 ms

The case of carpet

a2 = (0.42/0.021)×9.81 = 196 m/s²

t2 = 0.021×√(2/(0.42×9.81) = 0.015

s = 15 ms

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Answer:

FA < FB = FC

Explanation:

The buoyant force is equal to the weight of the displaced water.

The two that sink displace the same amount of water and therefore have the same buoyant force. The one that floats displaces less than its full volume of water and therefore has a lower buoyant force. Its buoyant force will equal its weight.

You do not have an answer option for the correct solution.

The quantity of heat energy that must be transferred to 2.25 kg of brass to raise its temperature from 20 °C to 240 °C is 195030 J

First, we shall list out the given parameters from the question. This is shown below:

The quantity of heat energy that must be transferred can be obtained as follow:

Q = MCΔT

= 2.25 × 394 × 220

= 195030 J

Thus, we can conclude quantity of heat energy that must be transferred is 195030 J

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To determine the spring constant of the scale, you can use the formula:

k = (F / x)

where k is the spring constant, F is the force acting on the spring, and x is the displacement of the spring.

In this case, the force acting on the spring is the weight of the water bottle, which can be calculated using the weight = mass * gravitational acceleration formula:

F = m * g

where m is the mass of the water bottle in kilograms and g is the gravitational acceleration of 9.8 m/s^2.

Plugging in the values given, we get:

k = (F / x) = (m * g / x)

= (350 g / (11.5 cm - 10.0 cm))

= 307.8 g/cm

This is the spring constant of the scale. It represents the force required to stretch the spring by 1 centimeter.

Answer:

10 Newtons.

Explanation:

F=MA

F=(5kg)(2 m/s^2)

10 N

Answer:the shape

Explanation:

Answer:

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

Answer:

2 Pascal (Pa)

Explanation:

Pressure is defined as the force acting per unit area. Mathematically, it is expressed as:

Pressure (P) = Force (F) / Area (A)

Given:

Force exerted by the air inside the balloon (F) = 1 N

Area of the balloon (A) = 0.5 m^2

Plugging in the given values into the formula for pressure, we get:

P = F / A

P = 1 N / 0.5 m^2

Using basic arithmetic, we can calculate the pressure inside the balloon:

P = 2 N/m^2

So, the pressure inside the balloon is 2 N/m^2, which is also commonly referred to as 2 Pascal (Pa) since 1 Pascal is equal to 1 N/m^2.

Answer:

1)   d = h- L - mg / k , 2)        k = 2 mg/h    (-1 +2 L / h)

Explanation:

1) Let's use the translational equilibrium equation

       \(F_{e}\) -W = 0

       F_{e} = W

       k x = mg

       x = mg / k

from the statement of the exercise the height of the bridge, for the reference system in the river, is

        h = L + x + d

        x = h - L - d

we substitute

        h - L -d = mg / k

        d = h- L - mg / k

2) They ask us for the spring constant. For this part we can use energy conservation

Starting point. At the point before jumping

        Em₀ = U = m g h

Final Point. When it's hanging

        Em_f =   \(K_{e}\) + U = ½ k x² + mg d

        Em₀ = Em_f

         mg h = 1 / k x² + m g d

in the exercise they indicate that Kate touches the surface of the river, so the distance d = 0

         mg h = ½ k x²

         k = 2 mg h / x²

we substitute the value of x

        k = 2mg  h / (h -L)²

        k = 2mg  h / (- L + h)²

we simplify the expression

       k = 2mg  h / [h² (1- L / h)²]

       k = 2m /h      (1- L / h)⁻²

In these jumps the bridge height is always greater than the length of the rope L / h <1, so we can expand the last expression

          (1- L / h)⁻² = 1 - 2 (1 -L / h) + 2 3/2!   (1 -L / h)² + ...

for simplicity let's keep up to the linear term, we substitute in the solution

       k = 2 mg/h    [1 - 2 (1- L / h)]

       k = 2 mg/h    (-1 +2 L / h)

Answer:

the average reaction time is 0.25 seconds.

Answer:

Positive, Negative

Explanation:

The image I've attached shows that vectors point into the negative source and vectors point away from the positive source.

Answer:

The truck will undergo the largest change in momentum if it has a greater mass than the small car.

Explanation:

The change in momentum of an object can be calculated using the equation:

Δp = m * Δv

where Δp represents the change in momentum, m represents the mass of the object, and Δv represents the change in velocity.

Since we are comparing the change in momentum of the car and the truck, we need to consider the masses of both vehicles.

Let's assume the mass of the car is represented by m_car, and the mass of the truck is represented by m_truck.

Since both vehicles collide head-on, the change in velocity (Δv) will be the difference between their initial velocities, considering that they are moving in opposite directions:

Δv = v_truck - v_car

Now, let's compare the change in momentum for the car and the truck:

For the car:

Δp_car = m_car * Δv

For the truck:

Δp_truck = m_truck * Δv

Comparing the magnitudes of the change in momentum, we can neglect the negative sign:

|Δp_car| = |m_car * Δv|

|Δp_truck| = |m_truck * Δv|

Since both Δv and Δp are positive values, we can conclude that the vehicle with the greater mass will undergo the largest change in its momentum.

Therefore, if the mass of the truck (m_truck) is greater than the mass of the car (m_car), then the truck will undergo the largest change in its momentum. Conversely, if the mass of the car is greater, then the car will undergo the largest change in its momentum.

Below are the required answers and explanations for each of the scenarios listed.

1. A machine that pulls all thermal energy out of a refrigerated space: This violates the second law of thermodynamics. This is because the second law of thermodynamics states that no heat engine can have an efficiency of 100 percent, and no heat transfer can occur from a colder to a warmer object without external work being done.

2. A machine that can pull 1000J of heat out of a refrigerated space and into a warmer space without external work: This violates the second law of thermodynamics. The second law of thermodynamics states that no heat transfer can occur from a colder to a warmer object without external work being done.

3. A machine that can turn 1000J of heat directly into 1000J of electricity: This does not violate any of the laws of thermodynamics.

4. A machine that can create 1000J of heat from 100J of electricity: This does not violate any of the laws of thermodynamics.

5. A machine that can pull 1000J of heat out of a refrigerated space and put 1500J of heat into a warmer space if it uses 500J of external work: This does not violate any of the laws of thermodynamics.

a) Option 2 is correct answer.

b) Option  2 is correct answer.

c) Option 4 is correct answer.

d) Option 4 is correct answer

e) Option 4 is correct answer.

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