Please answer :>
40 POINTS

Answer:

The first scenario!

Explanation:

W=F*d

a) 55*8= 440J

b) 60*6= 360J

c) 50*5= 250J

d) 40*10= 400J

The scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = F × d

Where F is force applied and d is distance travelled.

From the question;

A force of 55 N moves it 8 m

W = 55N×8m = 440N.m

A force of 60 N moves it 6 m.

W = 60N×6m = 360N.m

A force of 50 N moves it 5 m.

W = 50N×5m = 250N

A force of 40 N moves it 10 m.

W = 40N×10m = 400N.m

Therefore, the scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.

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Answer:

Chemical.

Explanation:

Weathering can be defined as the physical and chemical breakdown of rock into smaller pieces called sediment. Weathering can be classified into two categories namely;

I. Physical weathering : it is the process of breaking rocks into pieces without affecting its chemical composition e.g temperature, abrasion and frost action.

II. Chemical weathering : it is the process of breaking rocks into pieces by chemical action which leads to changes in its chemical composition e.g carbonation, hydration, plant acid and oxidation.

In this scenario, Lichen grows on the surface of hornblende granite and causes it to breakdown at a faster rate. Thus, this is an example of chemical weathering

Answer:

The value 0.51

Explanation:

(a) The original charge on the 40-pF capacitor is [tex]2 .0 \ \times \ 10^{-8} \ C[/tex].

(b) The charge on each capacitor after the connection is made is [tex]4 .0 \ \times \ 10^{-9} \ C[/tex].

(c)  The potential difference across the plates of each capacitor after the connection is 100 V and 400 V.

The original charge on the 40-pF capacitor is calculated as follows;

[tex]Q = CV\\\\Q = 40 \times 10^{-12} \times 500\\\\Q = 2 .0 \ \times \ 10^{-8} \ C[/tex]

[tex]C = \frac{C_1C_2}{C_1 + C_2} \\\\C = \frac{10 \times 10^{-12} \times 40 \times 10^{-12}}{10\times 10^{-12} \ + \ 40 \times 10^{-12}} \\\\C = 8 \times 10^{-12} \ F[/tex]

[tex]Q = Q_1 = Q_2\\\\Q = 8 \times 10^{-12} \ \times \ 500\\\\Q = 4 \times 10^{-9} \ C[/tex]

The potential difference across the plates of each capacitor after the connection is calculated as follows;

[tex]V = \frac{Q}{C} \\\\V_1 = \frac{Q}{C_1} \\\\V_1 = \frac{4 \times 10^{-9}}{40 \times 10^{-12}} \\\\V_1 = 100 \ V\\\\V_2 = \frac{Q}{C_2} \\\\V_2 = \frac{4 \times 10^{-9}}{10 \times 10^{-12} } \\\\V_2 = 400 \ V[/tex]

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Answer:

Gravitational potential energy is the greatest

Explanation

It is the highest point of a roller coaster.

Answer:

105 kg*m/s

Explanation:

p=mv

momentum=mass*velocity

momentum=15kg*7m/s

momentum=105kg*m/s

since the car moves, the force needed to move is greater than the frictional forces opposing it

a = 3.17m/s²

The acceleration of the car with a mass of 1300 kg which starts from rest generating a force of 4400 N and having a frictional force of 280 N is 3.17 m/s²

Force (F) = 4400 N

Frictional force (Fբ) = 280 N

Fₙ = F – Fբ

Fₙ = 4400 – 280

Thus, the net force acting on the car is 4120 N

Mass (m) = 1300 Kg

Net force (F) = 4120 N

Force = Mass × Acceleration

4120 = 1300 × a

Divide both side by 1300

[tex]a = \frac{4120}{1300}\\\\[/tex]

Therefore, the acceleration of the car is 3.17 m/s².

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Answer:

Explanation:

answer C looks good

there isn't really a "centrifugal " force. :/    when we are pushed "back" in a car seat.. it's not because there is a force pushing us backwards... but a force pushing us forwards.. just like when turning too,   a force pushes us into the corner,  not a force pushing out of the corner.  :)  

Answer:

The ans is 1 not 2

Explanation:

The distance between two adjacent node in a stationary wave is half wavelength π/2 and not full wavelength as in π,so when solving we got 2cm as the wavelength since they ask the dist. Also adjacent nodes you will divide it by 2 and the ans will give u 1cm.

Hope this helps.

Answer:

67 kg.

Explanation:

From the question,

Total momentum = (mass of the pilot+ mass of the helicopter)×velocity.

I = (m+M)v............................... Equation 1

Where I = Total momentum of the pilot and the helicopter, m = mass of the pilot, M = mass of the helicopter, v = velocity.

make m the subject of the equation

m = (I/v)-M.............................. Equation 2

Given: I = 7200 kg.m/s, v = 60 m/s, M = 53 kg.

Substitute these values into equation 2

m = (7200/60)-53

m = 120-53

m = 67 kg.

Hence the mass of the pilot is 67 kg

Answer:

why does anyone need to know

Explanation:

1 plus one = 69

This question is incomplete, the complete question is;

Time Dilation. A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 13.0 months . The scientists need a new laboratory to properly conduct measurements on the isotope. If laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 21.0 months to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00 x 10⁸ m/s.

If they place their sample of the isotope on the spaceship, at what speed must it travel in order for the new laboratory to be completed on Earth by the time half of the isotope on the spaceship decays? Assume that it took one month for the scientists to actually start construction and launch the spaceship so half the sample will remain in 12.0 months but it will still take 21.0 months to build the new lab.

Answer:

the required speed is 2.4618 x 10⁸ m/s

Explanation:

Given the data in the question;

Time taken to complete a lab = 21 months  

it took them one one to actually start so

Time remaining so that sample only decays by 1/2 = (13 - 1) = 12 months

we need to find at what speed the spaceship should travel so that time on earth ( = 12 months) become time on spaceship ( = 21 months)

we make use of time dilation equation;

t' = t/√( 1 - v²/c² )

where t is time in rest frame = 12 and t' is time in moving frame = 21

so we substitute

21 = 12/√( 1 - v²/c² )

21√( 1 - v²/c² ) = 12

√( 1 - v²/c² ) = 12/21

we square both side

( 1 - v²/c² ) = ( 12/21 )²

( 1 - v²/c² ) = 0.3265

v²/c² = 1 - 0.3265

v²/c² = 0.6735

v² = 0.6735 × c²

v = √0.6735 × √c²

v = 0.8206 × c

given that speed of light c = 3 x 10⁸ m/s

v = 0.8206 × 3 x 10⁸ m/s

v = 2.4618 x 10⁸ m/s

Therefore, the required speed is 2.4618 x 10⁸ m/s

Answer:

MgS

Explanation:

Answer:

465 grams of sugar.

Explanation:

I'm not sure this is true: is the relationship linear?

If it is then 300 grams of water should be able told 3*155 grams of sugar

3 * 155 = 465  grams of sugar in 300 grams of water.

Answer:

a. 3.73 m/s b. 27.8 m/s²

Explanation:

(a) Calculate his velocity (in m/s) when he leaves the floor.

Using the conservation of energy principles,

Potential energy gained by basketball player = kinetic energy loss of basket ball player

So, ΔU + ΔK = 0

ΔU = -ΔK

mg(h' - h) = -1/2m(v'² - v²)

g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor

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

9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)

9.8 m/s²(0.710 m) = -1/2(-v²)

6.958 m²/s² = v²/2

v² = 2 × 6.958 m²/s²

v² = 13.916 m²/s²

v = √(13.916 m²/s²)

v = 3.73 m/s

(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m

So, making, a subject of the formula, we have

a = (v² - u²)/2s

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

a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)

a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)

a = 13.913 m²/s²/(0.50 m)

a = 27.83 m/s²

a ≅ 27.8 m/s²

Answer:

The paper focuses on the biology of stress and resilience and their biomarkers in humans from the system science perspective. A stressor pushes the physiological system away from its baseline state toward a lower utility state. The physiological system may return toward the original state in one attractor basin but may be shifted to a state in another, lower utility attractor basin. While some physiological changes induced by stressors may benefit health, there is often a chronic wear and tear cost due to implementing changes to enable the return of the system to its baseline state and maintain itself in the high utility baseline attractor basin following repeated perturbations. This cost, also called allostatic load, is the utility reduction associated with both a change in state and with alterations in the attractor basin that affect system responses following future perturbations. This added cost can increase the time course of the return to baseline or the likelihood of moving into a different attractor basin following a perturbation. Opposite to this is the system's resilience which influences its ability to return to the high utility attractor basin following a perturbation by increasing the likelihood and/or speed of returning to the baseline state following a stressor. This review paper is a qualitative systematic review; it covers areas most relevant for moving the stress and resilience field forward from a more quantitative and neuroscientific perspective.

Explanation:

During a normal reaction to a stressful event, muscles are moved to their maximum capacity, and sensitivity is increased.

This is defined as the ability of an organism to respond to stimuli such as touch, sensation etc.

During exercise, sensitivity to substances such as insulin when glucose transport wears off helps to balance energy supply.

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Answer:

P=126W

Explanation:

Sorry if im wrong!

Answer:

125.5 watts

Explanation:

P=work/time

work=F*d

P=(F*d)/t

P=(15*251)/30

P=125.5 watts

Answer:

B. holding a coffee mug

Explanation:

Something must move a distance for work to be done.

work=force×distance this is the formula

Answer:

in my opinion i would say that at home cares of more effective because half of the time your child is safe and sound but in child care programs sometimes you have no idea what is going on when you are not around

Explanation:

Answer:

The Current decreases

Explanation:

HOPE THIS HELPS!

Answer:

C

Explanation:

sin of a right triangle is opposite divided by hypotenuse

in this case, that is 5/8.6 = 0.581395349

you can easily remember the trig functions with this acronym:

SOH-CAH-TOA

S: Sine

O: Opposite

H: Hypotenuse

C: Cosine

A: Adjacent

H: Hypotenuse

T: Tangent

O: Opposite

A: Adjacent

Explanation:

sin = Mp /0P

sin = 5/8,6

sin = 0,58

Answer:

Temperature

Explanation:

Answer:

Gravity

Explanation:

Gravity has nothing to do with weather. Temperature means hot or cold, water is rain and thunder, and wind is if it's in the winter or something like that

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

You don't have a image attached

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

          T² = ([tex]\frac{4\pi ^2}{ G M_s}[/tex] a³ = K_s a³

Let's apply this equation to our case

          a = [tex]\sqrt[3]{ \frac{T^2}{K_s} }[/tex]

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

             10² = K_s  [tex]a_{w}^3[/tex]

             a_w = [tex]\sqrt[3]{ \frac{100}{ K_s} }[/tex]

             a_w = [tex]\frac{1}{ \sqrt[3]{K_s} }[/tex]  4.64

Planet X

             a_x = [tex]\sqrt[3]{ \frac{640^3}{K_s} }[/tex]

             a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

              a_y = [tex]\sqrt[3]{ \frac{80^2}{K_s} }[/tex]

              a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

              a_z = [tex]\sqrt[3]{ \frac{270^2}{K_s} }[/tex]

              a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law