What is the Gravitational pull of something that weighs 254.4 Newtons and is at a 40 Degree angle?

If the metal box displaces water if 320 lbs, its weight in underwater is 250 lbs.

Archimedes principle states that when an object is totally or partially immersed in a fluid, it experiences un upthrust which is equal to the weight of the fluid displaced.

If the metal box displaces some volume of the water, it implies that the density of the metal box is greater than that of the water.

The weight of the metal box underwater is calculated as follows;

W = weight on land - weight of water displaced

W = 570 lbs - 320 lbs

W = 250 lbs

Thus, we can conclude that the weight of the metal box underwater is 250 lbs.

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Elements have atoms and protons which makes them the building blocks of matters.

What are elements?

The nature of elements make them impossible to broken down into simpler kinds of matter.

Elements can be grouped into:

In summary, elements have atomic number, structures, densities, and unique properties.

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The slope and displacement at point b. ei is constant. By using the method of virtual work we get Triangle B=6637.5 / EI.

The study of forces and movement in a mechanical system results in virtual work when the concept of least action is applied. A force's work on a particle as it moves along a displacement varies depending on the displacement. One will minimise the action among all virtual displacements, or potential displacements, that a particle might take. As a result, the particle follows this displacement in accordance with the principle of least action. The term "virtual work" refers to the labour performed by a force on a particle along a fictitious displacement.

virtual work and the associated calculus of variations were initially created to examine systems of rigid bodies, but they were later developed for the study of mechanics.

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If the ratio of the mass of star A to the mass of star B is 6 then, mass of star A is 22.3 solar masses and the mass of star B is 3.7 solar masses.

What is star?

The brilliant plasma spheroid that constitutes an astronomical object known as a star is held together by gravity. The star that is closest to Earth is the Sun. Even while other stars are also visible at night with the unaided eye, they appear as stationary points of light due to their immense distances from Earth.

A:B = 6:1

We can write it as

A/B = 6

A=6B

and we also given in the question as:

A+B = 26

putting the above value of A

6B+B = 26

B = 26/7

B = 3.7 solar masses

Subtracting it from total mass:

A = 26-B = 26-3.7

A = 22.3 solar masses

Therefore, mass of star A is 22.3 solar masses and mass of star B is 3.7 solar masses.

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Answer:

sleep well

bukas nalng yan kakapagod ehh

Answer:

1. Competition and prediction

Commensalism

Parasitism

Mutualism

Amensalism

2. As the population of wolves has declined the population of the moose increase

3.

The driver's reaction time during uniform motion is 1 second or 3 second.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

vo = 72 km/h = 20 m/s

vt = 0 m/s

a = -10 m/s²

sactual = 35 m

Find the instant distance without reacting

vt² = vo² + 2a . s

0² = 20² - 20s

s = 20 m

Find the reaction distance

sreact = sactual - s

sreact = 35 - 20

sreact = 15 m

Find the time reaction

sreact = vo . t + 1/2 . a . t²

15 = 20t - 5t²

t² - 4t +3 = 0

(t - 3)(t - 1) = 0

t = 3 s or t = 1 s

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The minimum amplitude of the wave is equal to  [tex]\frac{g \mu}{T_{\mathrm{s}}}\left(\frac{\lambda}{2 \pi}\right)^2[/tex]

A type of radio propagation known as a surface wave is known as ground wave propagation. Low and medium frequency versions of these waves travel across the surface of the earth. Mostly, these are employed for transmission between the earth's surface and the ionosphere. The quantity of individual waves makes up these.

The expression for the velocity of the wave is as follows:

[tex]v=\sqrt{\frac{T_s}{\mu}}[/tex]

Here,

Ts is the tension and

μ is the mass per unit length.

The expression for the wave vector is as follows:

[tex]k=\frac{2 \pi}{\lambda}[/tex]

λ is the wavelength of the wave.

The expression for the angular frequency is as follows:

ω = v k

Here, v is the velocity of the wave.

The expression for the minimum amplitude is as follows:

[tex]A=\frac{g}{\omega^2}[/tex]

Here, g is the acceleration due to gravity.

putting the values of v and k in w

[tex]\begin{aligned}&\omega=v k \\&\omega=\left(\sqrt{\frac{T_s}{\mu}}\right)\left(\frac{2 \pi}{\lambda}\right)\end{aligned}[/tex]

[tex]\begin{aligned}A=& \frac{g}{\omega^2} \\A=& \frac{g}{\left(\left(\sqrt{\frac{T_s}{\mu}}\right)\left(\frac{2 \pi}{\lambda}\right)\right)^2} \\&=\frac{g}{\frac{T_s}{\mu}\left(\frac{2 \pi}{\lambda}\right)^2} \\&=\frac{g \mu}{T_s}\left(\frac{\lambda}{2 \pi}\right)^2\end{aligned}[/tex]

The minimum amplitude of the wave is equal to [tex]\frac{g \mu}{T_{\mathrm{s}}}\left(\frac{\lambda}{2 \pi}\right)^2[/tex]

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Option (b) mgh/t will be the power supplied by the lifting force.

What does Newton's first law of motion state?

According to Newton's first law of motion, if a body is at rest or moving in a straight line at a constant speed, it will continue to move at that speed or remain at rest until acted on by a force.

How are formulas of power and force interlinked?

F = mg is the formula for force where,

F = force required

m = mass of the object

g = gravitational constant

and P=Fd/t is the formula for power where,

P= power

F=force

d= distance

and t= time

Substituting the value F=mg, in P=Fd/t we get,

P=mgd/t or P=mgh/t

Hence, we can conclude,

F = mg is the equation that expresses the force required to raise an item at a constant speed. P = Fd/t = mgh/t is then used to calculate the power.

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Following are the differences between substance and mixture

A material is said to be pure if it just contains that substance and cannot be broken down into any other components. It is possible to separate a mixture into two or more pure substances. Pure substances have distinct physical and chemical properties, whereas mixtures have various qualities depending on the pure substance amounts and placement within each mixture.

Pure substances might be molecules made up of two or more different elements, or they can be elements composed solely of one type of atom. Depending on how thoroughly the components are combined, mixtures can either be homogeneous or heterogeneous.

Following are the differences between element and compound

Compounds are chemical unions of two or more elements and can potentially be pure substances, whereas elements are invariably pure things. Real-world substances, such as elements and compounds, are rarely pure because they are frequently tainted during manufacturing, storage, or by their environment. In theory, pure denotes the absence of any discernible imperfections. The majority of the time, it is achievable to purify substances to that standard, albeit it frequently necessitates a lot of work.

Following are the differences between heterogeneous and homogeneous mixtures.

Mixtures can be physically separated since they contain multiple types of atoms and molecules. When a combination is homogenous, the particles are so small that the entire mixture appears to be formed of the same substance. When a mixture is heterogeneous, the qualities of the mixture vary depending on which section of the mixture is being analyzed. Typical homogeneous mixtures include solutions. For instance, a salt-water solution is a homogenous combination since it is possible to separate the water and salt using distillation to create pure water and crystalline salt.

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The statement that is most likely true about these area A is it has more food resources than area B.

In the given graph, it is clear that carrying capacity of area A is larger than that of area B. Larger the carrying capacity, larger will be its quantity of food resources.

Carrying capacity is the maximum population size that a habitat can support as per its available resources such as food, space etc. It can also be called as a species' average population size in a particular habitat.

The given question is incomplete. The complete question is in the link mentioned below.

Therefore, the statement that is most likely true about these area A is it has more food resources than area B.

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Answer:

Answer is a

Explanation:

Answer:

l-element. ll-mixture

lll-compound

Explanation:

element-2 or more substance make a element

mixture-2 or more substant is combined

compound-all the same and compact

Cage 2 will reach the bottom first. The time taken by cage one is 114.76 seconds and cage 2 is 0.32 seconds.

Given,

Mine depth Y = 2072 m

Cage speed 1 v = 65.0 km/h

Acceleration of Cage 2,

a = 4.00 102 m/s2

Solution:

We shall utilize the international system of measurements (SI) to interchange magnitudes using the unit system, which allows us to do so precisely and without the need for antiquities.

  v = 65 km / h (1000m/1Km) (1hr/3600s)

     = 18.06 m / s

The relationships between location, velocity, and acceleration can be discovered using kinematics.

Work each cage separately in this practice since there are two of them.

Cage 1

They suggest that it is moving at a constant speed, which we can determine using the laws of uniform motion.

v = d/t

v = d/v

t1 = 2072/18.06

t₁ = 114.76 s

Cage 2

Stop so that your starting velocity is zero.

y = v₀ t + ½ a t²

y = 0 + ½ a t²

[tex]t = \sqrt{2y/a} \\\\t2 = \sqrt{2*2072/4*10^{2} } \\[/tex]

t₂ = 0.32 s

On comparing the time taken by cage 1 and cage 2, we can say that cage 2 will reach the ground first.

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The acceleration in meters per second squared and in multiples of g is

a = 9.18 g.  

Acceleration is the rate of change of velocity with respect to time.

"comes to a stop" means

final v = 0 m/s

⇒v=0m/s.

"from an initial velocity of"

u = 0.600 m/s

"in a distance of 2.00 mm"

x  = 2.00 × (10)⁻³m

So, we need an equation that involves

a, v, u,  and x,

the equation is —

v²=(u)²-2×a×s

0= (0.600)² – 2×a× 2.00 ×(10)⁻³​

In solving this, we get a;

a=90 m/s² —>  eq. 1

Now, find the acceleration in meters per second squared and in multiples of g.

And, we know that g=9.8 m/s²2. —> eq.2

Divide eq.1 by eq.2

We get,

= 9.18 ⇒ a = 9.18 g.  

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The current in the 3.0 Ω resistor is mathematically given as

I_X=1.8A

This is further explained below.

The proportion of individual resistance to the overall resistance is the same as the current flowing through a branch to the current flowing through the whole system.

When the total current in a parallel circuit is already known, a shortcut approach for finding the branch currents in the circuit may be found via the use of a formula called the current divider formula.

In series circuits, the voltage drops across each resistor, but the current stays the same.

This is analogous to what happens in parallel circuits when the current splits between two branches.

The following diagram illustrates how the entire voltage is distributed across each resistor while maintaining the same amount of current throughout. Ohm's law states that V = I*R.

current dividers ruleis used when current is divided in parallel, such as it is in this case so we employ the current dividers rule as follows

Generally,  the equation for current dividers rule is mathematically given as

[tex]I_X=\frac{R_T}{R_X+R_T} I_T[/tex]

where,

I_X = current

R_X = resistor

R_T = total resistance

I_T = sum of the current flowing into a network where R _X is connected in parallel with R_T

Therefore

[tex]I_X=\frac{9}{3+9} 2.4[/tex]

I_X=1.8A

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The amount of time needed for a radioactively decaying substance to drop to half its initial amount is known as the half-life. 75% of the sample of thorium-227 decays in 38 days.

The half-life, in radioactivity, is the amount of time needed for half of the atomic nuclei in a radioactive sample to spontaneously transform into other nuclear species while emitting particles and energy, or, alternatively, the amount of time needed for the radioactive sample to disintegrate on average once every second.

Calculate the  number of half lives that must elapse to obtain amount needed. Let x the  no. of half life that elapse

Amount of thorium-227 remaining =

Original amount of thorium 227*1/2 for each half-life

1/4 = 1*1/2x

x = 2

The no. of days are required by  multiplying the half-life with the number of half-lives elapsed:

19/1 half-life*number of half-life = total days

19/1*2 = 38 days

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The area of the ball that weighs 95 N and exerting a pressure of 43 Pa is 1.51 m².

Area can be defined as the region bounded by a plane shape.

To calculate the area of the ball, we use the formula below.

Formula:

Where

Make A the subject of the equation

Froim the question,

Given:

Substitute these values  into equation 2

Hence, the area of the ball that weighs 95 N and exerting a pressure of 43 Pa is 1.51 m².

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Answer:W hen the car suddenly stops due to either braking or crashing, inertia will force the passengers or any other objects to keep moving forward. And since the force of inertia increases with the speed, the motion can varied from a light erk to a more serious crash.

Explanation:

tempreture

√12 is  degrees Celsius per meter.

What is tempreture?

tempreture is the degree of hotness or coldness of any kind of objects.

Sol-

The highest rate of tempreture increase is given as 6degrees as per the question

Imagine this high rate as a diagonal vector pointing in the northeast direction.

This vector has an x component (horizontal, or east) and a y component (vertical, or north).

The highest rate (maximum) is found as the magnitude of these two components.

It is the square root of (x^2)+(y^2).

6= √((x^2)+(y^2))

-> Take the square of both sides to remove the square root on the right side of the equation.

36 = (x^2)+(y^2)

-->The x^2 and y^2 values each must equal 8 in order to get:  12

(x^2) = 12

x = √12

(y^2) = 12

y = √12

If the object moves directly north, it is therefore moving only in the y-direction.

Thus, it is increasing tempreture at a rate of √12 degrees Celsius per meter.

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1) The control group are the people that are in group B

2) The independent  variable is the administration of the breath mint

3) The dependent variable is the breath of the participants

4) The conclusion of Mr Krab should be that the breath mint is effective in treating bad breath.

5)  10 people in group B reported fresher breath because they thought that they got the breath mint.

The term experiment is used to describe the way tat we could be able to spot the relationship between the dependent and the independent variable. Let us note that the independent variable is the variable that we have to manipulate in the course of an experiment while the dependent variable is the variable that would change as we change the independent variable.

We can see that there was a sort of improvement of the breath of the customers which according to the hypothesis of the experiment would be attributable to the new formula of the  breath mint.

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answer:

explanation: I don't know the answer to be honest

The mass of the rock is, m = 24 kg

The velocity of the rock is, v = 2 m/s

The kinetic energy of the rock can be given by the formula

Substituting the values, the kinetic energy will be

The relative velocity of the two monkeys is 8.568m/s.it overtakes and grabs onto a 6.20-kg monkey also moving east on a second vine. the first monkey is moving at 10.2 m/s at the instant it grabs the second.

The relative velocity of an object to another item is its speed in respect to that object. It is a measurement of the relative speed of two moving objects. Because it clarifies how objects move and interact with one another, relative velocity is significant in physics.

Given;

mass of first monkey = 5.2 kg

mass of second monkey = 6.2 kg

speed of first monkey = 10.2 m/s

speed of second monkey = 7.20 m/s

by conservation of momentum

initial momentum = final momentum

5.2 * 10.2 + 6.2 * 7.2 = (5.2 + 6.2) * v

v = 8.568 m/s

their common speed = 8.568 m/s

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Answer:

Time to reach zero speed       t1 = 65 ft/sec / 32 ft/sec^2 = 2.03 sec

H = 65 * 2.03 - 32/2 ^ 2.03^2 = 66 ft above launch point

h = distance ball falls = 66 + 3 = 69 ft

h = 1/2 g t2^2        time to fall 69 ft

t = (2 * 69 / 32)^1/2 = 2.08 sec

Total time = t1 + t2 = 2.03 + 2.08 = 4.11 sec

The velocity of the bullet when it left the gun is 118 m/s.

We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule

vt = vo + a . t

vt² = vo² + 2a . s

s = vo . t + 1/2 . a . t²

where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.

From the question above, we know that

vt = 118 m/s

g = 9.8 m/s²

Find the time taken to maximum height where the initial velocity is zero

vt = vo + at

118 = 0 + g . t

118 = 9.8.t

t = 12.04 s

Find the velocity of bullet when it left the gun

vt = vo - gt

0 = vo - 9.8 . 12.04

vo = 118 m/s

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A 90 kg astronaut, initially at rest, throws a 5 kg oxygen tank at 1 m/s away from himself.  The new speed for astronauts is 0.055 meters per second, and they will go in the opposite direction of the thing they are chasing.

This is further explained below.

Generally, For this, use the law of conservation of momentum

Total final momentum = Total final momentum

P_{a i}+P_{0 i}=P_{a f}+P_{o f}

m_{a v_{a i}}+m_0 v_{o i}=m_{a v_{a f}}+m_0 v_af

0+0=90 v_{a f}+5 *1

90 v_{a f}=-5 a

[tex]v_{a f}=-\frac{5}{90} \mathrm{~m} / \mathrm{sec} \\\\v_{a f}=-0.055 \mathrm{~m} / \mathrm{sec}[/tex]

In conclusion, The new speed of astronauts is  0.055m/second and astronauts will move in the opposite direction of the object.

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The magnitude of the acceleration when the crate's mass is 310 kg is 0.74 m/s² and the acceleration when the crate's weight is 310 N is 7.3 m/s².

What is acceleration?

The rate at which an item changes its velocity both in magnitude and direction is known as acceleration, a vector quantity. If an object's velocity is changing, it is accelerating.

Although the full specification of force,

[tex]F_{nett} = ma[/tex] where, m = mass, a = acceleration

This situation involves both x and y axes, only the x-application is needed to find what this particular problem asks for.

We note that acceleration in y axis is 0 so, we can denote acceleration in x axis simply as a. We choose +x to the right and +y up.  

Let's calculate the x component of the force working on the crate:

[tex]F_x = F cos38^0[/tex]

putting values from question, we get

Fx = 355 N

and the resistive force have magnitude (f) = 125 N

(a) As per the Newton's second law

Fx - f = ma

355 - 125 = 310 a

(acceleration) a = 0.74 m/s²

(b) In this case, By the equation:

mass = weight/gravitational acceleration

we get mass(m') = 310/9.8 = 31.6 kg

Now, by Newton's second law

Fx - f = m' a

355 - 125 = 31.6 a

(acceleration) a = 7.3 m/s²

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Complete question

A worker drags a crate across a factory floor by pulling on a rope tied to the crate. The worker exerts a force of magnitude 450 N on the rope, which is inclined at an upward angle 38° to the horizontal, and the floor exerts a horizontal force of magnitude 125 N that opposes the motion. Calculate the magnitude of the acceleration of the crate if (a) its mass is 310 kg and (b) its weight is 310 N.