100 POINTS. PLEASE PROVIDE EXPLANATION

Answer:

The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.

Explanation:

To convert from gram/centimeter³ to kilogram/meter³

First,

NOTE:

1 kilogram = 1000 gram, that is 1 gram = 0.001 kilogram; and

1 meter = 100 centimeter, that is 1 centimeter = 0.01 meter

then,

1 meter³ = 1000000 centimeter³, that is

1 centimeter³ = 0.000001 meter³

Now, we can write that

1 kilogram/meter³ = 1000 gram / 1000000 centimeter³

= 0.001 gram/centimeter³

If 1 kilogram/meter³ = 0.001 gram/centimeter³

Then, 1 gram/centimeter³ = 1/0.001 kilogram/meter³  = 1000 kilogram/meter³

Hence, 1 gram/centimeter³  = 1000 kilogram/meter³

∴The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.

That is,

2.33 gram/centimeter³ will be 2.33 × 1000 kilogram/meter³

= 2330 kilogram/meter³

Answer:

Low mass stars are: hydrogen burning in the core while on the Main Sequence. As the hydrogen fuel runs out, extreme pressure raises the temperature to 100 million degrees, where helium burning becomes possible.

Answer:

more

hope this helps

plz mark brainliest

Answer:

They are perpendicular.

Explanation:

Answer:

the last one

Explanation:

ii took the quiz and it was right... i think

Answer: Melting, evaporation and sublimation.

Melting, evaporation and sublimation require an increase in energy.

To determine the changes of state that require an increase in energy, we need to know about changes of state.

Melting, evaporation and sublimation are the changes of state of a substance.

Thus, we can conclude that melting, evaporation and sublimation require an increase in energy.

Learn more about changes of state here:

https://brainly.com/question/18372554

#SPJ2

Answer:

Negative

Explanation:

duh

Answer:

:)

Explanation:

negative.

go add the snap carmel.bratz

Answer:

The answer is Top-Down processing

Explanation:

I had this question on a apex quiz and i got it correct.

Answer: A system or frame of reference are those conventions used by an observer (usually standing at a point on the ground) to be able to measure the position and other physical magnitudes.

Answer:

C a position from which something is observed

om edu 2021

Explanation:

Answer:

A) v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B)√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) µ_s = tan θ

D) µ_s = 0.4663

Explanation:

A) The forces acting on the car will be;

Force due to friction; F_f

Force due to Gravity; F_g

Normal Force; F_n

Now, let us take the vertical direction to be j^ and the direction approaching the centre to be i^ downwards and parallel to the road surface by k^.

Also, we will assume that initially, F_n is in the negative k^ direction and that it will have a maximum possible value of; F_f = µ_s × F_n

Thus, sum of forces about the vertical j^ direction gives;

ΣF_j^ = F_n•cos θ − mg + F_f•sin θ = 0

Since F_f = µ_s × F_n ;

F_n•cos θ − mg + (µ_s × F_n × sin θ) =0

F_n = mg/[cos θ + (µ_s•sin θ)]

Also, sum of forces about the centre i^ direction gives;

ΣF_i^ = F_n(sin θ + (µ_s•cos θ)) = mv²/r

Plugging in formula for F_n gives;

ΣF_i^ = [mg/[cos θ + (µ_s•sin θ)]] × (sin θ + (µ_s•cos θ)) = mv²/r

Making v the subject gives;

v = √[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))]

B) What we got in a above is the minimum speed the car can have while going round the turn.

The maximum speed will be gotten by making the frictional force(F_f) to point in the positive k^ direction. This means that F_f will be negative.

Now, if we change the sign in front of F_f in the equation in part a that led to the minimum velocity, we will have the maximum as;

v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

Thus the range is;

√[(rg(tan θ - µ_s))/(1 + (µ_s•tan θ))] ≤ v ≤ v = √[(rg(tan θ + µ_s))/(1 - (µ_s•tan θ))]

C) For the minimum speed to be 0, it implies that F_f will be in the negative k^ direction. Thus, Sum of the forces in the k^ direction gives;

ΣF_k^ = mg(sin θ - µ_s•cos θ) = 0

Thus;

mg(sin θ - µ_s•cos θ) = 0

Making µ_s the subject gives;

µ_s = sin θ/cos θ

µ_s = tan θ

D) If θ = 25.0°;

Thus;

µ_s = tan 25

µ_s = 0.4663

Answer:

Explanation:

Answer:

v = - 1,715 m / s ,   x = 0.0156 m

Explanation:

This is an oscillatory movement exercise, which is described by the expression

          x = A cos (wt + Ф)

we can assume that the block is released from its maximum elongation, so the phase constant (Ф) is zero

As we are told that the stone does not affect the movement of the spring mass system, the amplitude and angular velocity do not change, in the upward movement the stone is attached to the mass, but in the downward movement the mass has an acceleration greater than g leave the stone behind, let's look for time, for this we use the definition of speed and acceleration

         v = dx / dt

         v = - A w sin wt

          a = - Aw² cos wt

         a = -g

         -g = - Aw² cos wt

          wt = cos⁻¹ (g / Aw²)

          t = 1 / w cos⁻¹ (g / Aw²)

angular velocity and frequency are related

        w = 2π f

        w = 2π 4

         w = 8π   rad / s

remember that the angles are in radians

          t = 1 / 8π cos⁻¹ (9.8 / (0.07 64π²))

          t = 0.039789 1.3473

          t = 0.0536 s

let's find the speed for this time

          v = - A w sin wt

          v = - 0.07 8π sin (8π 0.0536)

          v = - 1,715 m / s

the distance is

           x = A cos wt

           x = 0.07 cos (8π 0.0536)

           x = 0.0156 m

Answer:

Answer:

The object's mass.

Explanation:

The formula d=m/v.

d --> density

m--> mass

v --> volume

With density and volume given, you can calculate the mass of the object.

Explanation:

Answer:

Did a little research.

Explanation:

The Law of Action-Reaction:

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum). According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the accelerations of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club collides with a golf ball at rest upon a tee, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations. Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects that causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.

Consider the collision between a moving seven ball and an eight ball that is at rest in the sport of table pool. When the seven ball collides with the eight ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven ball experiences a leftward force that causes it to slow down; the eight ball experiences a rightward force that causes it to speed up. Since the two balls have equal masses, they will also experience equal accelerations. In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Answer:

A hypothesis is what you think will happen.

A conclusion is the results of an experiment summarized.

Hope this helps.

Answer:

it has to be b contact forces

Explanation: because the two objects are next to each other but they still have force

Answer:

c

Explanation:

Answer:

maybe the answer is in is D part

Answer: The control group is the part where you see what happens when you change a variable you want to study/examine. Basically, you need the control group because you need something to see what happens when you change something.

Hope this Helps! :))))

Answer:

Net force = 11340 N

Explanation:

Given that,

Mass of an automobile, m = 2100 kg

Acceleration of the automobile, a = 5.4 m/s²

We need to find the net force required for the automobile. The net force is the product of mass and acceleration. It can be given by the formula as follows :

[tex]F=ma\\\\F=2100\ kg\times 5.4\ m/s^2\\\\F=11340\ N[/tex]

So, the net force is 11340 N.

Answer:

i got you

Explanation:

Unfortunately, it is also an incomplete chemical equation. ... But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.

Answer:

28 m/s

Explanation:

vf^2=vi^2+2a(delta y)

=sqrt(2* -9.8 m/s^2* -40 m)

=28 m/s

Answer:

answer A is the correct one

Explanation:

Weber's law states that  "the smallest discernible change of a stimulus and proportional to the stimulus".

Applying this law to cases of optical intensity, the ratio must be

          k = cte = ΔI / I

where ΔI is the variation of the intensity and I is the value of the intensity

In general, for humans, the constant is 0.15 for the rods and 0.015 for the cones of the retina.

When reviewing the answers, answer A is the correct one, since in order for the previous relationship to be maintained, the magnitudes must rise proportionally

Answer:

152,000 N

Explanation:

(a) Initial potential energy = final kinetic energy

mgh = ½ mv²

v = √(2gh)

v = √(2 × 10 m/s² × 3.00 m)

v = 7.75 m/s

(b) Work done on pile hammer = change in energy

FΔy = 0 − (mgh + ½ mv²)

F (-0.074 m) = -((365 kg) (10 m/s²) (0.074 m) + ½ (365 kg) (7.75 m/s)²)

F (-0.074 m) = -11220.1 J

F ≈ 152,000 N

Answer:

The average kinetic energy

The amount of heat a substance has or the average kinetic energy of particles in a substance

Answer:

a) the astronauts and the station are in free fall towards the center of the Earth

Explanation:

Weightlessness is only a sensation of zero weight, a body experiences when it is in free fall towards the center of the Earth, caused by lack of contact force on such body.

Weightlessness doesn't mean the object has zero actual weight, is just a sensation of no weight due to lack of contact force to produce upward reaction on the object which gives the real sense of ones weight.

Thus, the astronauts experience apparent weightlessness because:

a) the astronauts and the station are in free fall towards the center of the Earth.

Answer:

a) [tex]\beta = 1.111\times 10^{-7}[/tex], b) [tex]\beta = 9\times 10^{-7}[/tex], c) [tex]\beta = 3.087\times 10^{-6}[/tex], d) [tex]\beta = 2.5\times 10^{-5}[/tex], e) [tex]\beta = 0.5[/tex], f) [tex]\beta = 0.877[/tex]

Explanation:

From relativist physics we know that [tex]c[/tex] is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) A car traveling 120 kilometers per hour:

At first we convert the car speed into meters per second:

[tex]v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)[/tex]

[tex]v = 33.333\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 33.333\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 1.111\times 10^{-7}[/tex]

b) A commercial jet airliner traveling 270 meters per second:

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 270\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 9\times 10^{-7}[/tex]

c) A supersonic airplane traveling Mach 2.7:

At first we get the speed of the supersonic airplane from Mach's formula:

[tex]v = Ma\cdot v_{s}[/tex]

Where:

[tex]Ma[/tex] - Mach number, dimensionless.

[tex]v_{s}[/tex] - Speed of sound in air, measured in meters per second.

If we know that [tex]Ma = 2.7[/tex] and [tex]v_{s} = 343\,\frac{m}{s}[/tex], then the speed of the supersonic airplane is:

[tex]v = 2.7\cdot \left(343\,\frac{m}{s} \right)[/tex]

[tex]v = 926.1\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 926.1\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 3.087\times 10^{-6}[/tex]

d) The space shuttle, travelling 27000 kilometers per hour:

At first we convert the space shuttle speed into meters per second:

[tex]v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)[/tex]

[tex]v = 7500\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 7500\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 2.5\times 10^{-5}[/tex]

e) An electron traveling 30 centimeters in 2 nanoseconds:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

[tex]v = \frac{d}{t}[/tex]

Where:

[tex]v[/tex] - Speed, measured in meters per second.

[tex]d[/tex] - Travelled distance, measured in meters.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]d = 0.3\,m[/tex] and [tex]t = 2\times 10^{-9}\,s[/tex], then speed of the electron is:

[tex]v = \frac{0.3\,m}{2\times 10^{-9}\,s}[/tex]

[tex]v = 1.50\times 10^{8}\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 1.5\times 10^{8}\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 0.5[/tex]

f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

[tex]v = \frac{d}{t}[/tex]

Where:

[tex]v[/tex] - Speed, measured in meters per second.

[tex]d[/tex] - Travelled distance, measured in meters.

[tex]t[/tex] - Time, measured in seconds.

If we know that [tex]d = 10^{-14}\,m[/tex] and [tex]t = 0.38\times 10^{-22}\,s[/tex], then speed of the electron is:

[tex]v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}[/tex]

[tex]v = 2.632\times 10^{8}\,\frac{m}{s}[/tex]

The ratio [tex]\beta[/tex] is now calculated: ([tex]v = 2.632\times 10^{8}\,\frac{m}{s}[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex])

[tex]\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }[/tex]

[tex]\beta = 0.877[/tex]

Answer:

A

Explanation:

Refraction