Weight of an object remains same on the Earth and the Moon.
true or false?

Reference point

Explanation:

I am not sure

Answer:

it represents the physical changes.

Explanation:

I

Answer:

226.8 kN/m

Explanation:

The work done by the spring, W equals the kinetic energy of the satellite, K

W = K

work done by the spring, W = 1/2kx² where k = force constant and x = extension of spring

kinetic energy of the satellite, K = 1/2mv² where m = mass of satellite = 1060 kg and v = speed of satellite = 3.35 m/s

1/2kx² = 1/2mv²

k = mv²/x²

Also, the spring force F = kx where k = force constant and x = extension of spring.

k = F/x

equation both expressions for k, we have

mv²/x² = F/x

x = mv²/F since F = ma where m = mass of satellite and a = maximum acceleration of satellite = 5.00g and g = 9.8 m/s²

x = mv²/ma = mv²/5.00mg = v²/5.00g

Substituting the values of the variables into the equation, we have

x = v²/5.00g

= (3.35 m/s)²/(5.00 × 9.8 m/s²)

= 11.2225 m²/s²/49 m/s²

=0.229 m

Now k = F/x = 5.00mg/x

Substituting the values of the variables into the equation, we have

k = 5.00mg/x

k = 5.00 × 1060 kg × 9.8 m/s²/0.229 m

k = 51940 kgm/s²/0.229 m

k = 51940 N/0.229 m

k = 226812.23 N/m

k = 226.81223 kN/m

k ≅ 226.8 kN/m

Answer:

[tex]v=11.135m/s[/tex]

Explanation:

From the question we are told that

Mass [tex]m=0.40kg[/tex]

Frictional friction [tex]f=1.2[/tex]

Spring constant [tex]k=800N/m[/tex]

Compressed [tex]x=0.50[/tex]

Generally the energy balance equation is mathematically given as

[tex]1/2kx^2=f*30+m*g*sin \theta*30+1/2m*v^2[/tex]

if [tex](sin=10/30-1/3)[/tex]

[tex]1/2(800)(0.5)^2=(1.2)*30+(0.4)*9.8*1/3*30+1/2(0.4)*v^2[/tex]

[tex]100=36+39.2+0.20v^2[/tex]

[tex]v^2=\frac{24.8}{0.20}[/tex]

[tex]v=\sqrt{124}[/tex]

[tex]v=11.135m/s[/tex]

Answer:

the average speed of the particles in terms of the speed of light is 0.81c

Explanation:

Given the data in the question;

[tex]t_p[/tex] = 256.2 ps = 256.2 × 10⁻¹² s

speed of light c = 2.99 × 10⁸ m/s

d = 10.7 cm = 0.107 m

we know that; Average speed v = d/t ------- let this be equation 1

Also, given that 256.2 ps is the lifetime of particle X frame, proper time will be;

t = Y[tex]t_p[/tex] = [tex]t_p[/tex] / √( 1 - [tex]\frac{v^2}{c^2}[/tex] ) --------- let this be equation 2

Next, we input equation 2 into equation 1'

v = d / [ [tex]t_p[/tex] / √( 1 - [tex]\frac{v^2}{c^2}[/tex] )  ]

v = d/[tex]t_p[/tex][ √( 1 - [tex]\frac{v^2}{c^2}[/tex] ) ]

v[tex]t_p[/tex]/d = √( 1 - [tex]\frac{v^2}{c^2}[/tex] )

we square both sides

( v[tex]t_p[/tex]/d )² = (√( 1 - [tex]\frac{v^2}{c^2}[/tex] ) )²

v²[tex]t_p[/tex]²/d² = 1 - [tex]\frac{v^2}{c^2}[/tex]

v²[tex]t_p[/tex]²/d² +  [tex]\frac{v^2}{c^2}[/tex] = 1

v²( [tex]t_p[/tex]²/d² +  [tex]\frac{1}{c^2}[/tex] ) = 1

v²( ([tex]t_p[/tex]²c² + d²)/d²c² ) = 1

∴

v² = (d²c²) / ([tex]t_p[/tex]²c² + d²)

v = √[ (d²c²) / ([tex]t_p[/tex]²c² + d²) ]

v = (dc) / √([tex]t_p[/tex]²c² + d²)

so we substitute

v = (0.107 m × c) / √( (256.2 × 10⁻¹² s)²(2.99 × 10⁸ m/s)² + (0.107 m )²)

v = 0.107c / √( 0.00586814 + 0.011449 )

v = 0.107c / √( 0.01731714 )

v = 0.107c / 0.1315946

v = 0.81c

Therefore, the average speed of the particles in terms of the speed of light is 0.81c

Answer:

(B) 80 ohms

Explanation:

Since the circuit is a series connection, simply add the two resistances together to get 80 ohms.

Answer:

the third one with the refrigerator

Answer:

12m/s

Explanation:

v=fλ

30×(40÷100)=

12m/s

Answer:

Explanation:

The correct answer is right

Answer:

True

Explanation:

Ton is a unit that is used to measure huge masses like cars, trucks and huge boxes

Answer:

Particles in agas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

Explanation:

hope this helps

Answer: sound waves travel faster in solid than water or air.

In vacuum sound waves don't travel at all

Answer:

A: Sound travels faster through air

Explanation:

The speed of sound is the distance travelled per unit of time by a sound wave as it propagates through an elastic medium. At 20 °C, the speed of sound in air is about 343 metres per second, or a kilometre in 2.9 s or a mile in 4.7 s.

Answer: it is the earth gravity :’)

Explanation:

Earth's gravity keeps the air molecules of the atmosphere from escaping the area surround Earth

We have that the  is mathematically given as

Fn=36000N

Question Parameters:

mass of 2.0 x 10^3 kg

the radius of curvature is 24 m and

the roller coaster car has a tangential speed of 17 m/s

Generally the equation for the Force   is mathematically given as

F=ma

Fn-fg=mv^2/r

Fn=m(v^2/r+g)

Therefore

Fn=2000(12^2/18+9.8)

Fn=36000N

For more information on Force visit

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Answer:

The answer is B since no one answered it :)

Explanation:

Answer:

b

Explanation:

i took the test

Answer:

B. The same on the moon.

Explanation:

The density of an object is the ratio of the mass contained by the object to the volume occupied by that mass.

[tex]Density = \frac{Mass}{Volume}[/tex]

When the object is taken from the earth to anywhere in the universe, its mass remains constant. The dimensions of the object and hence its volume also remains constant anywhere in the universe.

Therefore, the density of the object will also remain the same as it depends upon the mass and the volume of the object.

So, the correct option is:

B. The same on the moon.

Answer:

they can asked to sue

Explanation:

bud

any pics? so i can see the question ⁉️

Answer:

four-year degree Right arrow. graduate school Right arrow. residency

Explanation:

edge

Four-year degree Right arrow. graduate school Right arrow. residency. This career path would be best for Kieran to follow.

According to the Oxford English Dictionary, a person's "career" is their "course or progress through life." According to this definition, a person's "career" includes a variety of facets of their life, education, and employment. The term "career" is also widely used to refer to the professional parts of a person's life, as in "career woman," for instance. A profession or occupation that often requires specific training or formal education is referred to as a person's lifework in the third manner that the word "career" is employed.When used in this context, "a career" refers to a set of linked occupations that are often followed within a single industry or sector. For instance, one may speak of "a career in education," "a career in criminal justice," or "a career in the building trade."

To know more about career :

https://brainly.com/question/30040900

#SPJ3.

Answer:

they're reactions

Explanation:

The relationship between the valence electrons and position is: the number of valence electrons determines the position

This is the number of electrons in the outermost shell of an atom.

NOTE: The outermost shell is called valence shell

This is where an element is located in the periodic table

The position of an element in the periodic table is determined by the number of valence electrons.

For example

Sodium, Na (atomic number of 11) has the following electronic configuration

Since the valence electron is 1, thus, sodium is located in group 1 of the periodic table.

Thus, we can see that the position of an element in the periodic table is related to the valence electron(s) in the atomic shell of the element.

Learn more about valence electron:

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Answer:

The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90. Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

Answer:

6.15 s

Explanation:

From the question,

Applying the law of conservation of momentum

Momentum of the Heavier skater = Momentum of the lighter skater.

Mv = mV................. Equation 1

Where M = mass of the heavier skater, m = mass of the lighter skater, v = Velocity of heavier skater, V = velocity of the lighter skater.

But,

v = r/t........................ Equation 2

V = r/t'................ Equation 3

Where r = radius of the circular rink, t = time taken for the heavier skater to reach the edge, t' = time taken for the lighter skater to reach the edge.

Substitute equation 2 and equation 3 into equation 1

M(r/t) = m(r/t')............. Equation 4

Given: M = 65 kg, m = 40 kg, r = 50/2 = 25 m, t = 10 s.

Substitute into equation 4 and solve for t'

65(25/10) = 40(25/t')

162.5 = 1000/t'

t' = 1000/162.5

t' = 6.15 seconds

Answer:

the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Explanation:

Given that the data in the question;

angle of inclination with respect to the ground [tex]\theta[/tex] = 30°

length of plane d = 2m

m₁ = 1 kg

m₂ = 0.5 kg

now, velocity of the first block at midpoint;

[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

v² = gsin[tex]\theta[/tex]d

v = √( gsin[tex]\theta[/tex]d)

g is 9.8 m/s

so we substitute

v = √( 9.8 × sin30° × 2)

v = √( 19.6 )

v =  3.13 m/s

Now, velocity just after collision of the blocks will be;

(m₁ + m₂)v₂ = m₁v

v₂ = m₁v / (m₁ + m₂)

we substitute

v₂ = (1 × 3.13) / (1 + 0.5)

v₂ = 3.13 / 1.5

v₂ = 2.0866 m/s

now, final kinetic energy will be;

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²

we substitute

[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]

[tex]KE_f[/tex] = 7.35 + 3.2654

[tex]KE_f[/tex] = 10.62 J

Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Answer:

11.34 i think

Explanation:

Answer:

Relative Dating

Explanation:

Relative dating is used to determine a fossils approximate age by comparing it to similar rocks and fossils of known ages. Absolute dating is used to determine a precise age of a fossil by using radiometric dating to measure the decay of isotopes, either within the fossil or more often the rocks associated with it.