An athlete lifts a 330 N set of weights from ground level to a position over her head, a vertical distance of 2.05 m. Assume the weights are moved at constant speed. a) How much work does the athlete do? b) What is the mass of the weights?​

All of the following are examples of physical properties except tearing.

This is used to describe the state of a physical system and is usually measurable.

Examples include:

Tearing isn't an example of a physical property which was why option A was chosen as the most appropriate choice.

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The momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.

The momentum transfered to the more massive cart is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

⁻ΔP₁ = ΔP₂

ΔP₂ = m₂v₂ - m₂u₂

ΔP₂ = m₂(v₂ - u₂)

ΔP₂ = 3m₁(v₂ - u₂)

ΔP₂ = 3 x 3.8 x (1.7 - 0)

ΔP₂ = 19.38 kgm/s

Thus, the momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.

The complete question is beblow

A cart of mass 3.8 kg is traveling to the right (which we will take to be the positive x-direction for this problem) at a speed of 6.9 m/s. It collides with a stationary cart that is three times as massive. After the collision, the more massive cart is moving at a speed of 1.7 m/s, to the right.

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[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Heat capacity of body 1 :

[tex]\qquad \sf  \dashrightarrow \:m_1s_1[/tex]

Heat capacity of body 2 :

[tex]\qquad \sf  \dashrightarrow \:m_2s_2[/tex]

it's given that, the the head capacities of both the objects are equal. I.e

[tex]\qquad \sf  \dashrightarrow \:m_1s_1 = m_2s_2[/tex]

[tex]\qquad \sf  \dashrightarrow \:m_1 = \dfrac{m_2s_2}{s_1} [/tex]

Now, consider specific heat of composite body be s'

According to given relation :

[tex]\qquad \sf  \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2[/tex]

[tex]\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}[/tex]

[tex]\qquad \sf  \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }[/tex]

[ since, [tex] m_2s_2 = m_1s_1 [/tex] ]

[tex]\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}[/tex]

[tex]\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{ \cancel{m_2}(\frac{s_2}{s_1} + 1)}[/tex]

[tex]\qquad \sf  \dashrightarrow \:s' = \dfrac{ 2 s_2}{ (\frac{s_2 + s_1}{s_1} )}[/tex]

[tex]\qquad \sf  \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2} [/tex]

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Answer:

Fundamental quantities are quantities that were created to measure a object/substance which are the basis on which derived quantities are formed, where as derived quantities are created by extracting variables from the fundamental quantities.

Answer:

300N

Explanation:

in your drawing make sure to put 230n and 160n on the same side with arrows facing the same direction on the opposite side of the box add an arrow with 0n and an arrow to represent friction

Hello!

We know that at the BOTTOM of the pendulum's trajectory, the bob has a maximum speed. This means that its KINETIC ENERGY is at a maximum, while its Gravitational POTENTIAL ENERGY is at a minimum.

On the other hand, when the bob is at its highest points, the bob has a velocity of 0 m/s, so its KE is at a minimum and its PE is at a maximum.

We can use the work-energy theorem to solve. Let the Initial Energy equal the bob's energy at one of the sides, while the final Energy equals the bob's energy at the bottom.

[tex]E_i = E_f\\\\PE = KE[/tex]

Recall that:
PE = mgh

m = mass (kg)

g = acceleration due to gravity (m/s²)

h = height (m)

KE = 1/2mv²

m = mass (kg)

v = velocity (m/s)

Set the two equal and solve for 'h'.

[tex]mgh = \frac{1}{2}mv^2[/tex]

Cancel mass.

[tex]gh = \frac{1}{2}v^2[/tex]

Solve for 'h'.

[tex]h = \frac{v^2}{2g}\\\\h = \frac{3^2}{2(9.8)} = \boxed{0.459 m}[/tex]

Answer:

maybe line toward the finish

Explanation:that question is confusing

Velocity is the pace at which your displacement is shifting; it describes how quickly and in which direction you are travelling. And acceleration, which is measured in meters per second, is the rate at which your velocity is changing (or meters per second squared).

Net force and mass are the two primary factors that impact how quickly an item accelerates. For instance, mass has an inverse relationship with acceleration while net force has a direct relationship.

The acceleration is also influenced by other elements like friction, air or fluid resistance, and pressure.

According to the second law, the mass of the item and the net force acting on it both impact how quickly an object accelerates.

Therefore, an object's acceleration is directly proportional to the net force applied on it and inversely proportional to its mass.

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#SPJ2

Answer:

did you get it done if not lmk I will help you out tomorrow when I get up

Answer:

Speed of the ball when it was thrown = 16.67 m/s

Height reached by the ball = 14.16 s

Explanation:

Given time in the ball comes to the same height from where the ball was thrown is 3.4 s.

Let us assume that the ball was thrown with a speed of u in the upward direction. Since only gravitational force acts on the particle in the downward direction, it has a constant acceleration in the downward direction.

We also know that a particle has the same magnitude of velocity when it is under a free fall at a fixed height but direction is opposite i.e., initially in the upward direction and finally in the downward direction.

https://tex.z-dn.net/?f=%5Ctherefore%20%5Cdfrac%7B-u-u%7D%7B3.4%7D%3D%20-9.8%5C%5C%5CRightarrow%20%5Cdfrac%7B2u%7D%7B3.4%7D%3D%209.8%5C%5C%5CRightarrow%202u%20%3D%209.8%5Ctimes%203.4%5C%5C%5CRightarrow%20u%20%3D%20%5Cdfrac%7B9.8%5Ctimes%203.4%7D%7B2%7D%5C%5C%5CRightarrow%20u%20%3D%2016.67%5C%20m%2Fs

Hence, the particle was thrown with a speed of 16.67 m/s.

Now, we also note that the particle has zero velocity at its maximum height.

So, from the initial position to the maximum height reached by the particle, we have

https://tex.z-dn.net/?f=0%5E2%20%3D%20u%5E2%2B2as%5C%5C%5CRightarrow%20s%20%3D%20%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%5C%5C%5CRightarrow%20s%20%3D%20%5Cdfrac%7B0%5E2-16.67%5E2%7D%7B2(-9.8)%7D%5C%5C%5CRightarrow%20s%20%3D14.16%5C%20m

Hence, the maximum height reached by the ball is 14.16 m.

Answer: 20 m/s

Explanation:

Acceleration = change in velocity / time → 10 m/s^2 = final velocity - 0 (original velocity) / 2s → final velocity = 20 m/s

Answer:

Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.

Explanation:

Answer:

(A) 0.8 kgm/s

Explanation:

because of the even ground it would only slow down

Alpha

I hope this helps you

:)

Answer:

a

Explanation:

Distance is simply the distance travelled which in this case would be 4km + 3km = 7km

To work out displacement, try to imagine the situation.

Draw a straight line to the east (label it 3) and then draw another line from the end of the first line upwards (label this one 4). Thus, you've created a right angles triangle. Now use pythagorean theorem to work out the displacement

4^2 + 3^2 = 25

sqrt 25 = 5 = displacement

The volume of the material that must be used to make a star with the mass of the sun is 1.2×10⁵¹ m³.

Volume is the amount of space occupied by an object or a plane figure.

To calculate the volume of the material that must be used to make a star with the mass of the sun, we use the formula below.

Formula:

Where:

Make V the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, The volume of the material that must be used to make a star with the mass of the sun is 1.2×10⁵¹ m³.

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Answer:

1.     121.05 kg m/s NW

2.     94.5 kg m/s NW

3.     26.55 kg m/s NW

Explanation:

The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

The magnitude of the force each plate experiences due to the other plate is determined as follows;

F = U/d

where;

[tex]F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d} \\\\[/tex]

Q = CV

[tex]F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}[/tex]

where;

The capacitance is given as;

[tex]C = \frac{\varepsilon _o A }{d}[/tex]

[tex]F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}[/tex]

Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

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Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

when water freezes, the temperature, Фf = 0 °C

Q = 82 x 4200 x (0 - 0)

Q = 0

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

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Based on the data provided, the value of the constant b is 3.27 × 10^-5 kg/s and the time required to reach 63% of terminal velocity is 0.58 s.

The terminal velocity of a body is the velocity at which the body falls at constant velocity through a fluid.

For the falling raindrop, let positive direction be downwards and negative direction upwards,

The raindrop experiences a downward gravitational force mg, and an upward drag force -bv.

The total force at a time t is given as

a)

Terminal velocity is achieved then the total force is 0,

0 = mg - bv0

Therefore

b = mg/v0

Substitutingthe values:

b = (3 × 10^-5 × 9.8)/9

b = 3.27 × 10^-5 kg/s

b) Applying Newton's Second Law

F = ma

where

Therefore,

mg = mv/t t = v/g

however, t is at 63% velocity

thus:

t = 0.63v/g

t = 0.63 × 9 /9.8

t = 0.58 s

Therefore, the value of the constant b is 3.27 × 10^-5 kg/s and the time required to reach 63% of terminal velocity is 0.58 s

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Answer:25

Explanation: because higher means less kinetic energ

Answer:

m = 4.5 kg

Explanation:

w = - 1800 j

Fd = - 1800 j

mgd = - 1800 j

m = - 1800 ÷(gd)

m = - 1800 ÷( 10×-40)

m = 4.5 kg

Try this solution, all the details are in the attachment. note, the answer is marked with orange colour. If it is possible, check the provided solution in other sources.

Answer: ≈0.81c.

[tex] \tt{Difference \: Between \: \underline\red {Horticulture} \: And \: \underline\green{Agriculture}}[/tex]

[tex] \text{\underline\red {Horticulture}}[/tex]

The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.

[tex] \text{\underline\green{Agriculture}}[/tex]

It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.

Answer:

In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.

Answer:

I can only define a few of them, hope you don't mind.

Seismic- It means connected with or caused by earthquakes

Epicenter- It means the point on the earth's surface where the effects of an earthquake are most felt strongly.

Fault- A place where there is a break that is longer than usual in the layers of rock in the earth's crust.

Tectonic waves- Waves connected with the structure of the earth's surface.

Plate boundaries- The boundaries that differentiates the large sheets of rock (called PLATES) that form the earth's surface. Plate tectonics is the movement of the large sheets of rocks that form the earth's surface

[tex] \: \: \: \: \: \: \: \: \: [/tex]

Answer:

the process of bending of light when travling throw different mediume is called refraction of light.

Answer:

Explanation:

Use that:

[tex]F=-\vec{\nabla}U(x,y)=-\left(\hat{i} \frac{\partial U}{\partial x}+\hat{j}\frac{\partial U}{\partial y}\right)=-11.6x\hat{i}+10.8y^{2}\hat{j}[/tex]

Then use the 2nd Newton's Law of Motion:

[tex]\vec{a}=\frac{\vec{F}}{m}=\frac{-11.6x\hat{i}+10.8y^{2}\hat{j}}{0.04}=-290x\hat{i}+270y^{2}\hat{j}[/tex]

At x = 0.3 and y = 0.6, we can find the acceleration as:

[tex]\vec{a}=-87\hat{i}+97.2\hat{j}[/tex] (in SI unit)

Then the magnitude of the acceleration on that point is:

[tex]a=\sqrt{(-87)^{2}+(97.2)^{2}}\approx 130.44[/tex] (SI Unit)