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DochEvi [55]
2 years ago
10

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced
Physics
1 answer:
charle [14.2K]2 years ago
4 0

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

You might be interested in
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
gregori [183]

The answers are:

a) Work=125,923.61J

b) Power=1.46watt

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}

Also, we need to find the mass given the density of the blood.

1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg

Now, calculating how much work does the heart do in a day, we have:

Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J

Then, calculating what is the power output and its horsepower, we have:

Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt

Have a nice day!

7 0
2 years ago
In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat
Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

\frac{Q}{t} = \frac{kA\Delta T}{d}

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

\Delta T = The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

Q_2 = 2*Q_1

Replacing,

kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}

\frac{\Delta T}{x}=2*\frac{\Delta T}{x}

\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}

\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}

Solvinf for T_o,

T_o = -13.18

Therefore the temprature at the outside windows furface when the heat lost per second doubles is  -13.18°C

3 0
2 years ago
What are the reasons for the establishment of UGA?
3241004551 [841]

Answer:

In February 1784, just after the close of the Revolutionary War, the General Assembly of Georgia earmarked 40,000 acres of land to endow "a college or seminary of learning." The following year, Abraham Baldwin, a lawyer and minister educated at Yale University in New Haven, Connecticut, who had settled in Georgia

Explanation:

please mark this answer as brainliest

8 0
2 years ago
You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh
hjlf

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8  since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083  

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121  = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

4 0
2 years ago
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
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