What is the density of the paint if the mass of a tin containing 5000 cm3 paint is 7 kg. If the mass of the empty tin, including the lid is 0.5 kg.​

The potential energy of the climber is 6.94 × 10⁶. It is the energy which is present in the body of an object which is at rest. Thus, the correct option is A.

Potential energy is the energy which is present in the body of an object which is at rest. This energy is transformed into kinetic energy when the body experiences any force and undergo motion. The SI unit of potential energy is Joule (J).

PE = m × g × h

where, PE = Potential energy,

m = mass of the object,

g = acceleration due to gravity,

h = height

PE = 80 × 9.8 × 8848

PE = 80 × 9.8 × 8848

PE = 784 × 8848

PE = 6936832

PE = 6.94 × 10⁶

Therefore, the potential energy of the climber is 6.94 × 10⁶.

Therefore, the correct option is A.

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Answer:

C.

Explanation:

That is the main part of a cell, apart of the nervous system and controls all the other functions of the cell.

I hope this helped

Brainliest if it did? No pressure!

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Answer:

You have to compare the densities,

Explanation:

To find density:

1. Measure the object's weight

2. Divide the mass by the volume

Answer:

where is the picture of star

Answer:

cell body → dendrite → axon → axon terminals

Explanation:

im in edgunity

Answer:

Material that allow the electrons to move freely in order to produce a current

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Answer:

KE = 1/2 w I^2    where w is the angular frequency

V = w R   and w = 50 m/s / .33 m = 150 / sec

w = 150 / sec for wheel

Each spoke also has angular velocity of 150 / sec about the center of the wheel since it makes one revolution as the wheel makes one revolution

I1 = 1/3 m L^2   inertia of 1 spoke

i1 = 1/3 * .01 kg * m^2 / 9    = .00037 kg-m^2

I50 = .0185    inertia of 50 spokes about axle

I2 = M R^2 = 1.2 * (1/3)^2 = .1333 kg-m^2

I (wheel) = (.0185 + .1333) = .152 kg-m^2

E = 1/2 * 150 * .152^2 = 11.4 kg-m^2 / sec^2 = 11.4 Joules

1 Joule = M * a * s = kg * m/s^2 * m = kg-m^2 / s^2

The statement that describes the transformation of the graph of function f onto the graph of function g is: The graph shifts 2 units right and 7 units down.

To determine the transformation of the graph of function f onto the graph of function g, we compare the two functions f(x) and g(x) and observe the changes in the equations.

The function f(x) = (1/x - 3) + 1 represents a reciprocal function that is shifted vertically 1 unit up and horizontally 3 units to the right. The reciprocal function is reflected about the line y = x.

The function g(x) = (1/(1 + 4)) + 3 simplifies to g(x) = 4 + 3 = 7, which is a constant function representing a horizontal line at y = 7.

By comparing the equations, we can see that the transformation from f(x) to g(x) involves the following changes:

The term 1/x in f(x) is replaced by the constant 1/(1 + 4) in g(x), resulting in a vertical shift of 7 units up.

The term -3 in f(x) is replaced by 3 in g(x), resulting in a vertical shift of 3 units up.

The +1 in f(x) is replaced by +3 in g(x), resulting in an additional vertical shift of 2 units up.

Therefore, the overall transformation is a shift of 2 units to the right and 7 units down.

Hence, the correct statement is: The graph shifts 2 units right and 7 units down.

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Answer:

it is 3 times greater, because 30=3x10

Answer:

Units named after scientists are written in lowercase but their symbols are written as capital

Unit of power is watt, since it is named after a scientist, then symbol will be written as W

Farad, symbol = F

Henry, symbol = H

Units whose names aren't culled from that of scientists are written in lower case and their symbols are also in lower case.

Units such as meter, kilogram should be represented with symbols in small letters as: m and kg respectively.

Explanation:

Kindly check answer

The mass of the planet is  5.98 × 10^24 kg.

To calculate the mass of the planet, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit.

First, we need to convert the period of the satellite's orbit to seconds. We know that there are 60 minutes in an hour, so the period can be expressed as (6 × 60 + 12) minutes, which equals 372 minutes. Multiplying this by 60 seconds, we get a period of 22,320 seconds.

Next, we need to find the semi-major axis of the orbit. In a circular orbit, the semi-major axis is equal to the radius of the orbit. Therefore, the semi-major axis is 8.8 × 10^6 m.

Now, we can apply Kepler's Third Law to calculate the mass of the planet. The formula is T^2 = (4π^2/GM) × a^3, where T is the period of revolution, G is the gravitational constant, M is the mass of the planet, and a is the semi-major axis of the orbit.

Rearranging the formula, we can solve for the mass of the planet:
M = (4π^2/G) × a^3 / T^2

Plugging in the values, we get:
M = (4 × π^2 / 6.67430 × 10^-11) × (8.8 × 10^6)^3 / (22,320)^2

Evaluating this expression, we find that the mass of the planet is approximately 5.98 × 10^24 kg.

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Answer:

If you know the current and voltage across the whole circuit, you can find total resistance using Ohm's Law: R = V / I.

Explanation:

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\(\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]\)              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\(\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°\)

The distance r is:

\(r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m\)

You replace the values of all parameters in the equation (1):

\(\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}\)

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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Answer:

I KNOW THE ANSWER IT WILL COST 30$

Explanation:

(0.43 ± 0.08)m/s is the speed of the ball.

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. Speed can be thought of as the rate at which an object travels a distance, and it is a scalar quantity that denotes "how rapidly an object is traveling."

A fast-moving object travels at a high speed and completes a significant distance in a brief period of time. This is in contrast to an object traveling slowly, which travels a relatively short distance in the same length of time. Zero speed refers to an object that is not moving at all.

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The data does not support his hypothesis because the heaviest mass travelled the least distance.

The weight of an object is the force of gravity acting on the object and the force of gravity increases with increase in the mass of the object.

W = mg

where;

Based on law of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = ( v² ) / ( 2g )

The height travelled by an object increases with increase in speed of the object and heavier objects will have greater speed, hence travel greater height.

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The complete question is below:

Jacobie wanted to test the projectile motion of a pumpkin. His hypothesis was if the pumpkin had a larger mass, then the higher he could throw it, because the larger pumpkins would have more force. He bought three orange pumpkins and tested them all on the same day. His results are shown at right. Does the data support his hypothesis?

5kg = 10meters

10kg = 5 meters

15kg = 1 meter

Answer:

occipital parietal

Explanation:

Hope this helps, plus I took the test

Answer:

It is Conductivity because it is the measure of the ease.

The water flow rate from the faucet, assuming a steady flow, is approximately 0.0001608 cubic meters per second.

The phenomenon of a water stream "necking down" as it falls can be explained by the principle of conservation of mass. According to this principle, the flow rate of water remains constant throughout the system.

In the given figure, the water stream emerges from a faucet with an initial cross-sectional area of 1.8 cm² (Area one). As the water falls, it narrows down to a smaller cross-sectional area of 0.3 cm² (Area two) at a height of 25 cm.

To calculate the water flow rate assuming steady flow, we can use the formula:

Flow rate = A₁√(2gh) / A₁² - A₂²

Where A₁ and A₂ are the initial and final cross-sectional areas, respectively, g is the acceleration due to gravity, and h is the height of the water column.

Plugging in the given values, we have:

Flow rate = 1.8 cm² √(2 × 9.8 m/s² × 25 cm) / (1.8 cm²)² - (0.3 cm²)²

Converting the units to match, we get:

Flow rate = 0.018 m² √(2 × 9.8 m/s² × 0.25 m) / (0.018 m²)² - (0.003 m²)²

Simplifying the expression, we find:

Flow rate ≈ 0.0001608 m³/s

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Athletes training for the 100-meter dash will mostly rely on phosphocreatine for this event's rapid energy.

Numerous creatine kinases catalyze the reversible phosphorylation of creatine, which includes both the forward and backward reactions. It is used to detect tissue injury when creatine kinase (CK-MB, creatine kinase myocardial band) is present in blood plasma.

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Answer:

Explanation:

F = m*a = (0.5)*15 = 7.5 N

At 20 years old, the boy would require eyeglasses with lenses of approximately -1.49 diopters power in order to read a book at 25 cm.

1/f1 = 1/v - 1/u1

1/f1 = 1/∞ - 1/0.25 (converting 25 cm to meters)

1/f1 = 0 - 4

1/f1 = -4

f1 = -1/4

f1 = -0.25 meters

The initial lens power (P1) is the reciprocal of the focal length:

P1 = 1/f1

P1 = 1/-0.25

P1 = -4 diopters

Now let's calculate the final focal length (f2) using the final distance (v2) of 40 cm:

1/f2 = 1/v2 - 1/u1

1/f2 = 1/0.40 - 1/0.25

1/f2 = 2.5 - 4

1/f2 = -1.5

f2 = -1/1.5

f2 = -0.67 meters

The final lens power (P2) is the reciprocal of the focal length:

P2 = 1/f2

P2 = 1/-0.67

P2 ≈ -1.49 diopters

Therefore, at 20 years old, the boy would require eyeglasses with lenses of approximately -1.49 diopters power in order to read a book at 25 cm.

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To find the kinetic energy at position B, we need to know the height or the velocity at position B. Without this information, we cannot calculate the exact value of the kinetic energy.

To determine the kinetic energy of the ball at position B, we need to consider the m conservation of Mechanical energy. Since the ball is released from position A, we can assume that there is no initial kinetic energy (velocity is zero), and the total mechanical energy at position A is equal to the potential energy.

The potential energy at position A can be calculated using the formula:

Potential energy at A = mass * gravitational acceleration * height

Potential energy at A = 20 kg * 9.8 m/s² * height

Now, at position B, all the potential energy is converted into kinetic energy. The kinetic energy at position B is given by the formula:

Kinetic energy at B = 1/2 * mass * velocity²

Since the ball is released from rest, the velocity at position B can be determined using the conservation of mechanical energy:

Potential energy at A = Kinetic energy at B

20 kg * 9.8 m/s² * height = 1/2 * 20 kg * velocity²

Simplifying the equation, we get:

9.8 m/s² * height = 1/2 * velocity²

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Answer:

  B. When Ellen started pedaling her bike, she began moving along the street

Explanation:

Kinetic energy is energy of motion. Ellen's kinetic energy changed when she went from rest to being in motion (began moving).

Answer:

Total resistance in parallel is Rp

So (1/Rp) = (1/12)+(1/16)+(1/20)

Rp = ?

impedance is 25

25 = Rp + Rs

put the value of Rp you'll get the answer.

The force actually being used to push the mower along the ground is 191 N.

When the physics student exerts a force of 250 N along the handle of the push mower, it's important to consider the components of this force that contribute to the actual force used to push the mower along the ground.

To determine the force used to push the mower along the ground, we need to find the horizontal component of the applied force. The angle of 40° indicates that the applied force can be broken down into two components: the horizontal component and the vertical component. The vertical component of the force is perpendicular to the direction of motion and does not contribute to pushing the mower forward.

To find the horizontal component, we can use trigonometry. The horizontal component is given by the formula:

Horizontal component = Applied force * cos(angle)

Plugging in the values, we get:

Horizontal component = 250 N * cos(40°)

Calculating this value, we find that the horizontal component of the applied force is approximately 191 N.

Therefore, the force actually being used to push the mower along the ground is 191 N. This is the component of the applied force that contributes to the forward motion of the mower, while the remaining vertical component is directed perpendicular to the ground and does not assist in pushing the mower forward.

By exerting a force of 250 N along the handle at a 40° angle, the student effectively applies 191 N of force to push the mower along the ground, ensuring efficient use of their effort while considering the environmental impact.

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