Is the weight lifter doing the same amount of work when he lifts the 1200N barbell 1m the same as when he lifts the same barbell 2m? Explain answer


Is my answer correct in the first problem?

Answer:

μ = 0.55

Explanation:

Given:

m = 43 kg

a = 0.79 m/s²

F pull = 265 N

_____________

μ - ? Friction coefficient

2 Newton's law:

m·a = Fpull  - μ·m·g

μ·m·g = Fpull - m·a

Friction coefficient:

μ = (Fpull - m·a) / (m·g)

μ = (265 - 43·0.79) / (43·9.8) ≈ 0.55

1. The first formula for average velocity is v = (x2 - x1)/(t2 - t1), where v is the average velocity, x1 and x2 are the initial and final positions, and t1 and t2 are the initial and final times. This formula is used when the motion of an object is known to be in a straight line.

Average velocity is the average rate of change of an object's position over a period of time. It is calculated by dividing the total displacement by the total time taken during the displacement. Average velocity is the result of an object's motion and is the overall velocity of the object. Average velocity is a vector quantity, meaning it has both magnitude and direction. Average velocity can be calculated for any type of motion, including linear, circular, and projectile motion. Average velocity is different from instantaneous velocity, which is the velocity of an object at a specific point in time.

2. The second formula for average velocity is v = Δs/Δt, where v is the average velocity, Δs is the total displacement (the difference between the initial and final positions), and Δt is the total time. This formula is used when the motion of an object is not known to be in a straight line.

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Answer:

8.803 s

(but this is only if it's a free fall, which we assume since we didn't get the mass of the hulk)

Answer:

The weight and the mass of an object are related because the weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. w=mg

Kepler's laws of planetary motion made a significant breakthrough in understanding the orbits of celestial bodies better than previous models.

Kepler's laws of planetary motion were a significant improvement over previous models for explaining the orbits of celestial bodies. He identified three main breakthroughs that made his models more accurate than those of previous models. Kepler discovered that the orbits of planets are not circular, as had been assumed previously, but elliptical. The second breakthrough came in the form of a law that states that planets move more quickly when they are closer to the sun than when they are farther away. Finally, Kepler discovered that the time it takes for a planet to complete its orbit is proportional to the cube of its distance from the sun. These three laws of planetary motion are Kepler's main answer to explaining the orbits of celestial bodies better than previous models.

By using them, Kepler could predict the positions of planets with far greater accuracy than anyone else of his time. It was through these laws that Kepler arrived at a conclusion that the planets move in elliptical orbits around the sun.

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Explanation:

Acceleration is the rate of change of velocity with time. When acceleration increases a body moves a faster velocity.

A vehicle at time 100s will have a faster velocity compared to one at t = 20s

The following are true :

C. At point 2 the concentrations of H20...O2... & H2 are all changing toward their equilibrium concentrations

D. At point 1 more reactants are converted to products than products are converted to reaction.

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Answer:

true

Explanation:

Answer:

True.

Explanation:

Answer:

In a fusion reaction, two light nuclei merge to form a single heavier nucleus

Explanation:

The time needed for the bacteria to reach 50,000 is 8.8 hours.

A thousand bacteria that grow at a rate of 65.3% per hour.

The number of bacteria forms a geometric sequence.

aₙ = a₁ × (rⁿ⁻¹)

aₙ = a₁ × (rⁿ⁻¹)

50,000 = 1,000 × (1.653ⁿ⁻¹)

50,000 ÷ 1,000 = (1.653ⁿ⁻¹)

50 = (1.653ⁿ⁻¹)

n - 1 = log₁.₆₅₃ 50

n - 1 = (log₁₀ 50) ÷ (log₁₀ 1.653)

n - 1 = 1.699 ÷ 0.218

n - 1 = 7.8

n = 7.8 + 1

n = 8.8 hours

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Answer:

The static "frictional force" and the "kinetic friction" will act on the box.

When  A force is applied to a box to move it to the right across the kitchen floor, static friction force comes into play before start moving. When the box starts moving, dynamic friction force comes into play.

The resistance provided by surfaces in touch as they move past one another is known as friction.

To walk without slipping, traction is provided by friction. In most situations, friction is advantageous. They do, however, give a strong amount of opposition to the move.

When  A force is applied to a box to move it to the right across the kitchen floor, static friction force comes into play until it starts moving. When the box starts moving, dynamic friction force comes into play.

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The time necessary for the largest ball to reach a velocity 95% of the terminal velocity is 22.35 seconds, and for the smallest ball, it is 0.23 seconds.

Given data:

Diameter of the largest ball, d₁ = 9.52 mm

Diameter of the smallest ball, d₂ = 2.38 mm

Viscosity of glycerin, η = 1.0 Pa. s

The terminal velocity is given as:

vT = 2r²(ρs - ρ)f / 9η

where,

r is the radius of the ball,

ρ is the density of the ball,

ρs is the density of glycerin and

f is the acceleration due to gravity.

Substituting the values, we have

vT₁ = (9.52 / 2)²(ρ₁ - ρ)f / 9ηvT₂

= (2.38 / 2)²(ρ₂ - ρ)f / 9η

where, vT₁ is the terminal velocity of the largest ball and

vT₂ is the terminal velocity of the smallest ball.

The time necessary for each ball to reach a velocity 95% of the terminal velocity is given as:

t = (-m/η) ln(1 - 0.95(v/vT))

where, m is the mass of the ball,

v is the velocity of the ball and

vT is the terminal velocity of the ball.

Substituting the values, we have for the largest ball:

v = 0.95vT₁t = (-m/η) ln(1 - 0.95(0.95vT₁)/vT₁)

= 22.35s (approx)

for the smallest ball:

v = 0.95vT₂t = (-m/η) ln(1 - 0.95(0.95vT₂)/vT₂)

= 0.23s (approx)

Therefore, the time necessary for the largest ball to reach a velocity 95% of the terminal velocity is 22.35 seconds, and for the smallest ball, it is 0.23 seconds.

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The substance that forms due to a chemical reaction is product. The correct option is G.

A chemical reaction is a process that converts one or more substances, known as reactants, into one or more various substances, known as products.

Chemical elements or substances are examples of substances. A chemical reaction reconfigures the constituent atoms of the reactants to produce various products.

Living things (cells, animals, humans, and insects) use chemical reactions to grow, develop, reproduce, and adapt.

When chemical bonds between molecules are formed or broken, chemical reactions occur.

The substances that initiate a chemical reaction are known as reactants, and the substances that result from the reaction are known as products.

Thus, the correct option is G.

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Cart A has a mass of 2 kilograms and a speed of 3 meters per second. Cart B has a mass of 3 kilograms and a speed of 2 meters per second. Compared to the inertia and magnitude of momentum of cart A, cart B has (D) greater inertia and the same magnitude of momentum.

To understand this answer, let's first define the terms inertia and momentum.

Inertia: Inertia is the resistance of an object to change its state of motion. It is directly proportional to the object's mass. A greater mass means greater inertia.

Momentum: Momentum is the product of an object's mass and its velocity. Mathematically, it is expressed as p = mv, where p represents momentum, m is the mass, and v is the velocity.

Now, let's compare the inertia and momentum of Cart A and Cart B:

Cart A:
Mass (mA) = 2 kg
Velocity (vA) = 3 m/s

Cart B:
Mass (mB) = 3 kg
Velocity (vB) = 2 m/s

1. Inertia comparison:
Since Cart B has a greater mass (3 kg) than Cart A (2 kg), Cart B has greater inertia.

2. Momentum comparison:
Momentum of Cart A (pA) = mA * vA = 2 kg * 3 m/s = 6 kg m/s
Momentum of Cart B (pB) = mB * vB = 3 kg * 2 m/s = 6 kg m/s

Since both carts have the same magnitude of momentum (6 kg m/s), their momentum is equal.

In conclusion, compared to Cart A, Cart B has greater inertia and the same magnitude of momentum.

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Answer:

This slide shows the three forces that act on a baseball in flight. The forces are the weight, drag, and lift. Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion

Answer:

27.9617 mi/h

Explanation:

Speed= distance/time

S= 15000/1200

Answer:

Oxygen starts with six valance electrons and ends with six valance electrons but after bonding has parts of the densities of 8 electrons

Explanation:

Chemical bonding does not increase the valance electrons. Oxygen has 6 valence electrons. Bonding is sharing of valance electrons

The electron density of two of the valance electrons of Oxygen are shared with the Hydrogen atoms. The Hydrogen atoms is turn share part of the electron density of their single electron with Oxygen.

This means that there are parts of 8 electron densities around the Oxygen. This does not increase the number of valance electrons of Oxygen. (Note the sharing is very unfair, Oxygen gets more than its

share of the electron densities of the two electrons used for the chemical bond.

Explanation:

The statement that matches refraction is Waves bend while passing from one medium to another.

Generally, Your definitions for the terms reflection, diffraction, transmission, absorption, and refraction are correct.

Here are a few more examples to help solidify your understanding:

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According to Pythagoras's Theorem, the square of the hypotenuse side in a right-angled triangle is equal to the sum of the squares of the other two sides.

A, B, and C are the three positive integers, and the Pythagorean triples are written as a2+b2 = c2. These triples are shown as (a,b,c). Here, the right-angled triangle's base, hypotenuse, and perpendicular are denoted by letters a, b, and c, respectively. The most popular and diminutive triplets include (3,4,5).

O is at a right angle in AOB.

∴ By using the Pythagoras Theorem, (AB² = (OB)²+ (OA)²

AB² = 4² + 3²

AB = √(4² + 3²)

AB = 5

∴ He has traveled 5 kilometers.

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Answer:

The pound originally was 12 oz, but was increased to 15 and then 16 oz later. One should remember that the weights on the scales were marked in oz, of 1, 2, 4, 8, &c, and in one case, the 16 oz simply became the pound.

Explanation:

Hope this helps

Answer: Pa=Pv

Explanation:

Khan academy

Amos runs up a flight of stairs with a total change in energy ΔEA. Vanessa exerts double the amount of energy in twice the time.

How does Amos’s power output, PA, compare with Vanessa’s power output, Pv?

Answer: Pa=Pv

Answer:

hydrochlorine +12÷B to the power of 4 -× y reapeated zminus 2 to the power of 9

Given:

Pressure, P = 300 kPa

Volume, V = 100 cm³

Temperature, T = 30°C

Let's fin the pressure of the gas inside the container if it is heated to 100°C.

Apply the Gay-Lussac's law:

Where:

P1 = 300 kPa

T1 = 30 + 273 = 303 K

T2 = 100 + 273 = 373 K

V1 = V2 (since the container is sealed).

Let's solve for P2.

Rewrite the formula for P2:

Therefore, the pressure if the container is heated to 100°C is 369.31 kPa.

ANSWER:

369.31 kPa

The loudness of a sound is related to the logarithm of the ratio of the measured intensity, 1, to a reference intensity, I. The loudness, L, of a sound, is measured in decibels, dB, and can be determined using the formula L=10log10 (I0I).

Given, The intensity of the sound of a rocket launching = 4500 times that of a jet engine. The loudness of the rocket launching, L = 170 dBNow, we can determine the value of L0 as follows:L = 10 log10 (I0/I)170 = 10 log10 (I0/I) (Equation 1)Therefore, I0/I = antilog (17) (from Equation 1)I0/I = 50,119.41Since the loudness of the rocket launching, L = 170 dB is already given, we can calculate the loudness of the jet engine as follows:L = 10 log10 (I0/I)dB = 10 log10 (I0/I)dB = 10 log10 (50,119.41)dB = 10 (4.700)dB = 47

The intensity of a rocket launching sound is 4500 times that of a jet engine sound, and its loudness is already provided as 170 dB. The loudness of a sound is related to the logarithm of the ratio of the measured intensity to a reference intensity, I.

To calculate the loudness of a jet engine, we can use the formula L = 10 log10 (I0/I).To determine I0/I, we substitute the loudness of the rocket launching, 170 dB, into the formula. We find that I0/I is equal to 50,119.41. We then substitute this value into the formula for the loudness of the jet engine. We find that the loudness of the jet engine is 47 dB. To the nearest decibel, the loudness of the jet engine is 47 dB.

Therefore, the loudness of the jet engine is 47 dB. 150 words to calculate the loudness of a jet engine, we must first determine the intensity of the sound it produces.

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The value of  angle θ through which the mass moves through is 12.15⁰.

The acceleration of an object is the rate of change of velocity of the object with time.

The acceleration of the mass is calculated as follows;

a = v²/r

where;

a = (2.25²)/(2.4)

a = 2.11 m/s²

tan θ = a/g

where;

tan θ = 2.11 / 9.8

tan θ = 0.215

θ = tan⁻¹(0.215)

θ = 12.15⁰

Thus, the value of  angle θ through which the mass moves through is 12.15⁰.

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To calculate the wavelength of a radio station's frequency, we can use the formula:

Wavelength = Speed of Light / Frequency

The speed of light is approximately 3 x 10^8 meters per second.

Given that the frequency of WYEP radio station is 9.13 x 10^7 hertz (91.3 megahertz), we can plug in the values:

Wavelength = (3 x 10^8 m/s) / (9.13 x 10^7 Hz)

Dividing the values, we get:

Wavelength = 3.28 meters

Therefore, the wavelength of WYEP radio station's frequency is approximately 3.28 meters.\(\)\(\huge{\mathcal{\colorbox{black}{\textcolor{lime}{\textsf{I hope this helps !}}}}}\)

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Answer:

do you mean how? if so

Explanation:people used newspapersand still kinda do back then they had posters up as they still do and newspapers