A CROW BAR WITH LENGTH 200 CM IS USED TO LIFT A LOAD OF 600N . IF THE DISTANCE BETWEEN FULCRUM AND LOAD IS 0.75. CALCULATE ; a, effort b, MA c, VR

Answer:

Magnets have two poles, called north and south. The like poles are attracted to unlike poles, but like poles repel each other. For example, the north pole of one magnet is attracted to the south pole of another. ... To make objects move with a magnet attach a piece of metal, or another magnet, to it.

The mass of the beach ball is 0.33 kg

From the question,

We are to determine the mass of the beach ball.

Using the formula

ρ = mv

Where ρ is the momentum

m is the mass

and v is the velocity

From the given information,

ρ = 3 kg m/sec

v = 9 m/sec

Putting the parameters into the formula, we get

3 = m × 9

∴ m = 3 ÷ 9

m = 0.33 kg

Hence, the mass of the beach ball is 0.33 kg

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Answer:If the elevator accelerated downward then the tension force smallest then 500 N. Otherwise, if the elevator accelerated upward then the tension force larger then 500 N.

The tension force = 750 N because the elevator accelerated upward. Force acts upward has plus sign and force acts downward has minus sign.

T – w = m a

750 – 500 = 50 a

250 = 50 a

a = 250 / 50

a = 5.0 m s–2

Explanation:

Answer:

Synthesis reactions are chemical reactions where two elements combine to make a product. Decomposition reactions are chemical reactions where a reactant produces another product, usually two, or even more. This is when the bonds in the compound are broken apart to make new compounds.

Explanation:

Answer: B the material with the greatest specific heat

Explanation:

Just did the assignment

Answer:

B. The material with the greatest specific heat

Explanation:

Answer:

They are both moving at the same speed since they will have the same acceleration

F = M a    describes acceleration

M a = M g sin theta    where theta is the slope

One can easily see that

a = g sin theta   and the weight does not enter into the equation

Without friction, the car cannot stop...

Answer:

F = G M m / R^2 = 6.67E-11 * 1.99E30 * 5.98E24 / (1.5E11)^2

F = 6.67 * 1.99 * 5.98 / 2.25 E21 Newtons

F = 3.53E22 N

Answer:

Yes

Explanation:

Because putting them together would make a type of energy making the answer yes

The order of the location of each process in the carbon cycle are as follows;

Reasons:

Ingestion; During the ingestion process, the products of the direct Sunlight and photosynthesis, which

are energy rich chemical substances are ingested by herbivores.

Decomposition; The plants and animals die and undergo decomposition.

Fossilization; A long period the decomposed plants and animals manifest

into hard petrified materials, with their parts transformed to fossils.

Combustion; The remains of plants and animals, subjected to heat and

pressure due to compression, as when as bacteria decomposition that

removes several other elements, leaving a carbon and hydrogen rich

sludge that can be processed into fuel used for combustion. The

combustion process produces carbon dioxide which restarts the cycle.

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If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is

F' = G (5m₁) (2m₂) / d²

F' = 10  Gm₁m₂ / d²

but this is really just F' = 10F. So J is the correct choice.

Hi there!

Using Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{d^2}}[/tex]

Fg = Force of gravity (N)

G = Gravitational Constant

m₁ = mass of object 1 (kg)

m₂ = mass of object 2 (kg)

d = distance between the objects (m)

There is a direct relationship between the masses of the objects and the resulting force of gravity, so we can plug new values in:

[tex]F_g = G\frac{5m_12m_2}{d^2}} = 10G\frac{m_1m_2}{d^2}} = 10F_g[/tex]

Thus, the correct answer is J. 10Fg.

Answer:

Produce energy, or make carbohydrates or fats.

Explanation:

These Milky Way companion galaxies are easily visible from dark locations in the Southern Hemisphere. Prime examples of erratic galaxies are the Large and Small Magellanic clouds (left and right, respectively).

In comparison to a rich cluster, the poor cluster typically has a slightly more erratic shape. A number of smaller galaxies orbit each major spiral. The Small and Large Magellanic clouds are the two most well-known examples of atypical galaxies.

When two galaxies collide, irregular galaxies frequently result. This unusual Cartwheel Galaxy was created when a tiny galaxy slid through the centre of a massive spiral galaxy.

Therefore, Rich clusters are other clusters that include hundreds to thousands of galaxies. A weak cluster can't cling to its members strongly because of its low bulk.

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Answer:

0.667; 4.965

Explanation:

Look at the picture I attached for the force analysis.

a) The coefficient=Friction/Normal Force. Because it's at constant speed, the force of friction + mgsin58.1° (because it's on an inclined plane and has split forces) is equal to the applied force (70.4N). Normal force is not equal to weight force though, because the box is on an inclined plane; it's equal to mgcos58.1°.

b) If it were to break, then the box no longer has an applied force, and the direction of friction has changed to up the inclined plane. F=m/a, so acceleration = mgsin58.1°- Friction/mass

Answer:

Breh seriously. Ugh fine.

1.B

2.D

3.C

4.C

5.D,A and B

6.A,C and D

Newton's second equation of motion :-

S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]

This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.

So let's see what pieces of information (bundles of equations) do we have with us, initially.

We have, a very primary equation with us,

dS/dt = v… (I)

(Considering motion in a straight line only)

And we also have the equation

dv/dt = a…(II)

Simply replacing the v in eqn (II) by eqn (I), we find

d2S/dt^2 = a…(III)

This is what we need to solve. It's easy.

You know,

d2S/dt^2 = d/dt(dS/dt) = a

⟹ dS/dt = ∫adt = at+c1

Since, dS/dt is the velocity of the particle,

Therefore, at t = 0, dS/dt|t = 0 = u

⟹ u = a∗0 + c1 = c1

⟹ c1 = u

Therefore, dS/dt = u + at

Thus, S = ∫(udt + atdt)

⟹ S = ut + 1/2at^2 +c^2

If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0

This makes S = S0 +ut + 1/2at^2

Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),

We get

S = ut + 1/2at^2

Hope it helps :)

Answer:

cellular respiration.

explanation: cellular respiration, the process by which organisms combine oxygen with foodstuff molecules, diverting the chemical energy in these substances into life-sustaining activities and discarding, as waste products, carbon dioxide and water.

[tex]\\ \sf\Rrightarrow KE=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \sf\Rrightarrow 8=\dfrac{1}{2}v^2[/tex]

[tex]\\ \sf\Rrightarrow v^2=16[/tex]

[tex]\\ \sf\Rrightarrow v=√16[/tex]

[tex]\\ \sf\Rrightarrow v=4m/s[/tex]

Answer:

the answer is 4m/s

[tex]\text{Given that,}\\\\\text{Velocity, v= 311 m s}^{-1}\\\\\text{Displacement, s = 9000 m}\\\\\text{Since velocity is constant,}\\\\s=vt\\\\\implies t =\dfrac{s}v = \dfrac{9000}{311} = 28.94 ~ \text{sec}.[/tex]

Answer:

he expected a fire to happen I think

Answer:

When Ernest Rutherford aimed a beam of particles at a thin gold foil he expected that the particles would be deflected slightly after passing through the foil.

Explanation:

hope it helped

Answer: 60 times brighter

Answer:

Sample answer: part 1:

What was your hypothesis?

This is what you wanted to prove or study with the experiment. Reading the question I can see that the experiment seems to study the relation between the viscosity of honey and the temperature, so the hypothesis may be something like:

"The viscosity of honey decreases as temperature increases".

According to your data, do you think your hypothesis was correct?

Here you need to see your experimental data, probably, you found that yes, as you increase the temperature the viscosity decreases.

The last thing we need to do here is to summarize difficulties and problems that you had in the experiment, and propose how you could solve them in order to have a better experiment.

Obviously, this depends on how you performed the experiment, but for example, if you saw that the temperature of the honey increased too fast, you could say:

"one problem was that the temperature of the honey increased too fast"

And one way to solve that would be submerging the honey in a water bath so the temperature increases a bit slower.

part 2:

Viscosity depends strongly on temperature, but this relation depends also on the given material.

Generally, we would see that the viscosity decays exponentially as the temperature increases.

So for the first question:

What effect did the temperature have on the viscosity of the honey?

You should have seen that, as the temperature increased, the viscosity decreased.

Now, we need to give some practical examples where knowledge of viscosity is important.

The first one is for cleaning: We know that if we increase the temperature, the viscosity decreases. Thus if we have some object with some viscous thing on it, increasing the temperature will cause it to be easier to clean.

Another example is to store/transport viscous things, for example, honey, it is actually a lot easier to pour honey in bottles if you first heat it a little, so it becomes less viscous.

So there you have two examples where knowing about viscosity (and how it relates to temperature) may be important.

Explanation:

Answer:

F1 = G M m / R^2       current force on earth

F2 = G M * 2 m / R^2     force if mass were doubled

F2 / F1 = 2         the force would be twice as large

The gravitational force between two objects not only depends on the mass but also the distance between them. If the distance to the star is same as to sun, then by doubling mass force also doubles. Thus option C is correct.

Gravitational force is the attractive force by which an object attracts other objects into its centre of mass. We are all standing in the ground because of the gravitational pull by earth.

The gravitational force between two objects of mass M1 and M2 at a distance r is written as follows:

g = G M1 M2/ r ² . where, G is gravitational constant.

As per this relation, gravitational force is directly proportional to the mass and inversely proportional to the square of the distance.

If the star has a distance from earth equal to the distance between sun and earth, but its mass is double the mass of sun, then the gravitational force will be two times compared with the sun.

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The magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

Given that a racquetball with a mass of 42 g is moving with a horizontal speed of 7 m/s to the right (+x direction).

mass m  = 42g = 42/1000 = 0.042kg

initial velocity before collision u = 7 m/s

It hits the wall of the court and rebounds to the hitter with a horizontal speed of 7m/s to the left (-x direction). That is,

velocity after collision v = 7 m/s

To calculate the magnitude of the racquetball's change in momentum, we will use the formula below

Change in momentum = Mv - Mu

Since momentum is a vector quantity, we will consider the direction.

Change in momentum = 0.042 x 7 - ( 0.042 x - 7)

Change in momentum = 0.294 + 0.294

Change in momentum = 0.588 kgm/s

Therefore, the magnitude of the racquetball's change in momentum is 0.59 kgm/s approximately.

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A) Figure attached below

B) The linear speed of the coin = 0.59 m/s

C) Linear speed as coin begins to slip = 0.83 m/s

D) The tangential speed will remain the same as seen in part C

Given data :

mass of coin = 0.0050 kg

Distance of coin from the center of disk = 0.14 m

Time to make a complete revolution = 1.5 s

A) Diagram showing the vectors on the figure is attached below

Linear speed of coin = 2 * π * ( 0.14 ) / 1.5

                                   = 0.59 m/s

given that: friction between coin and disk = 0.50

Friction becomes maximum when coin begins to slip

Maximum frictional force (Fmax) =  uV

where V = mg

Equating equations ( 1 ) and ( 2 ) to determine the linear speed ( v )

v² = u*r*g

∴ v = √(u*r*g ) = √( 0.5 * 0.14 * 9.8 )

                        = 0.83 m/s

D) If the experiment is repeated with a second coin glued to the top of the first coin the tangential speed will remain the same

Hence we can conclude that The linear speed of the coin = 0.59 m/s Linear speed as coin begins to slip = 0.83 m/s , The tangential speed will remain the same as seen in part C

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Answer: Electromagnetism

The term which is defined by the interaction between two charged particles is known as electromagnetism.

The study of charge and the fields and forces it generates is known as electromagnetism. Electromagnetism has two components: electricity and magnetism.

The special theory of relativity, developed by Albert Einstein in 1905, proved beyond a shadow of a doubt that both are components of the same reality. But in reality, magnetic and electric forces behave very differently and are modeled by various equations. Electric charges can produce forces whether they are stationary or moving. On the other hand, magnetic forces are only created by moving charges and only affect charges that are moving.

Thus, the term is known as electromagnetism.

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Answer:

I think b sorry if wrong

Explanation:

it b sorry if wrong

Answer:

Given that these particles make up atoms, they are often referred to as subatomic particles. There are three subatomic particles: protons, neutrons and electrons. Two of the subatomic particles have electrical charges: protons have a positive charge while electrons have a negative charge:

Answer:

2.1 seconds

Explanation:

let D represent the height in meters

let g represent gravity in meters per second squared

let t represent time in seconds

D= -42 meters (negative because it falls down)

g= -9.8m/s^2

t= [tex]\frac{-42 meters}{9.8 m/s^2}[/tex]

this gives t= 4.2857 s^2

the meter units cancel out in the division

now we must simplify the units to singular seconds so...

[tex]\sqrt{4.3 s^{2} }[/tex] which simplifies to [tex]\sqrt{4.3}[/tex] x [tex]\sqrt{s^{2} }[/tex]

this equals 2.1 seconds