016 (part 1 of 2) 10.0 pointa objects with masses of 232 kg and 638 kg are separated by 0.342 m. a 60.6 kg mass is placed midway between them. 232 kg 538 kg 60.6 kg © - 0.342 m find the magnitude of the net gravitational force exerted by the two larger masses on the 60.6 kg mass. the value of the universal gravi tational constant is 6.672 × 10-11 n • m? /kg? answer in units of n. 017 (part 3 of 2) 10.0 points leaving the distance between the 232 kg and the 638 kg masses fixed, at what distance from the 538 kg mass (other than infinitely remote ones) does the 60.6 kg mass experience a net force of zero? answer in units of m

The velocity of the jet is calculated to be 900 miles/hr when the relation between velocities of both propeller and the jet is given.

The distance travelled by a propeller plane and a jet = 3000 miles

Velocity of the plane Vp = 1/3 Velocity of the jet Vj

Vp = 1/3 Vj

Vj = 3 Vp ----(1)

Time taken by the propeller plane to complete the trip = 10 hours

Velocity of the propeller plane = Distance of the propeller plane/Time

⇒ 3000/10 = 300 miles/hr

The velocity of the jet = 3 times the velocity of the propeller plane

⇒ 3 × 300 miles/hr = 900 miles/hr

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The net acceleration working on the rock is 56.9 m/s² in upward direction, if the mass of the rock is 7.45 kg and the force applied on it is 497 N.

The upward force applied on the rock, F = 497 N

Mass of the rock, M = 7.45 kg

Acceleration due to gravity, g = 9.81 m/s²

We know force = Mass × Acceleration

Acceleration produced in the rock due to the force applied, a = 497/7.45

= 66.71 m/s²

Acceleration due to gravity is working downward, so net acceleration working on the rock, = 66.71 - 9.81 = 56.9 m/s²

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The charge on the larger of the two-point charges is 90 coulombs

To determine the charge on the larger point charge, we can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the known values of the force (120 N), distance (0.5 m), and the universal constant (8.99 x 10^9 Nm^2/C^2), we can use the equation F = k(q1q2)/r^2 to solve for one of the charges.

By rearranging the equation and substituting the known values, we get q1 = (Fr^2)/k. Then by substituting the given values, we get q1 = (120 N * (0.5 m)^2) / (8.99 x 10^9 Nm^2/C^2) = 90 coulombs.

Since we know the total charge is 180 coulombs and q1 is 90 coulombs, then the charge on the second point charge would also be 90 coulombs.

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The minimum number of degrees  is  108°.

Based on the accompanying figure,

The angle of rotation from point A to point B is the smallest angle at which the star can overlap itself.

Let a° be the angle formed by A and B.

By connecting the star's vertices, we can form a pentagon.

The formula to calculate a polygon's internal angle is

The internal angle measurement is

=> 180 x (n-2)/n

=> 180 x (5-2)/5

=> 180 x 3/5

=> 36

ABC measured as:

= 108°

Since the inner angle is divided into three equal parts by a side star,

36° is the internal angle measurement.

the AOB,

The sum of a triangle's inner angles is 180 degrees, or m(OAB) + m(AOB) + m(ABO).

36° + a + 36° = 180°

a + 72° = 180°

a = 180 - 72

a = 108°

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The full question is :

What is the minimum angle of rotation (in degrees) that will carry the star onto itself?

If the string is at the breaking point at the bottom of the circle, then the speed of the object at this point is 7.82m/s.

The force transferred through a rope, string, or wire when pushed by forces acting from opposite sides is referred to as tension. The tension force is applied along the entire length of the wire and exerts an equal amount of strain on the bodies at each end. Every physical object that comes into contact with another one applies some sort of force.

At the bottom of the circle,

Let the maximum speed will be V.

It is given when the string breaks its tension exceeds 22.5N

it means, T=22.5N

now, we know,

In a Circular path,

⇒ [tex]\frac{mV^2}{r} = T-mg[/tex]

⇒ [tex]V^{2}=\frac{r}{m}(T-mg)[/tex]

⇒ [tex]V^{2}=r(\frac{T}{m} -g)[/tex]

⇒ [tex]V=\sqrt{r(\frac{T}{m} -g)}[/tex]

According to the question,

m = 0.275KG, r = 0.850m and T = 22.5N

Putting the values, We get,

⇒ [tex]V=\sqrt{(0.850)(\frac{22.5}{0.275}-9.80 )}[/tex]

⇒ [tex]V=7.82m/s[/tex]

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The complete question is:

A 0.275 Kg ball wings in a vertical circular path on a string L=0.850m long. If the string breaks when its tension exceeds 22.5N, what is the maximum speed the ball can have at the bottom before that happens?

The work done on a block of certain mass when pushed with specific force is -162 J.

Mass of the block = 3 kg

Force with which the block is pushed = 12 N

Distance pushed = 9 m

The net force is nothing but the force minus the frictional force on the block.

Mathematically, Fnet = F - mg = 12 - (3×10) = -18 N

The work done is given by the formula, W = F × d

where, F is the force

d is the distance

W is the work done

Plugging in the values in equation, we have,

W = F × d = -18 × 9 = -162 J

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Iron production was mastered by Zhou craftsmen.

The Zhou dynasty was one of the greatest in Chinese history, and among its many achievements were the artisans who discovered how to make iron. Iron became a common material for agricultural equipment, which was revolutionary for the Chinese people and extremely beneficial for agriculture. Everything became lot easier, quicker, and more efficient as a result.

The Zhou dynasty, which followed the Shang dynasty, was the longest to rule China.

Artisans are individuals who create handicrafts.

Agriculture, education, military planning, Chinese literature, music, philosophical schools of thought, social stratification, as well as political and religious advancements, all benefited from the Zhou Dynasty's considerable cultural contributions.

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The net charge on the pith ball in coulombs when the number of electrons is specified is calculated to be 2.96 × 10⁻² C.

Given that, number of electrons = 1.85 × 10¹⁷

We know the charge on electron as, e = 1.6 × 10⁻¹⁹ C

The relation between charge and number of electrons is known to be,

Q = n e

where, Q is charge

n is number of electrons

e is charge on electron

After putting the values into the equation above, we have,

Q = 1.85 × 10¹⁷ × 1.6 × 10⁻¹⁹ = 2.96 × 10⁻² C

Thus, the net charge is calculated to be 2.96 × 10⁻² C.

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The estimated temperature in the stellar atmosphere is T = 4.48 / 8.62 x 10^-5 * ln([H2]/4) .

To estimate the temperature of a stellar atmosphere consisting almost entirely of hydrogen, we can use the concept of dissociation equilibrium. When a gas is in equilibrium, the rate of formation of molecules from atoms is equal to the rate of dissociation of molecules into atoms. The dissociation equilibrium constant (Kd) is given by the ratio of the concentration of atoms to the concentration of molecules, at a given temperature and pressure.

In this case, it is given that 50% of the hydrogen molecules are dissociated into atoms, so the concentration of atoms is half the concentration of molecules. The pressure is also given as 100 Pa.

The dissociation equilibrium constant (Kd) can be calculated using the following equation:

Kd = [H]^2 / [H2]

Where [H] is the concentration of atomic hydrogen and [H2] is the concentration of molecular hydrogen.

The equilibrium constant can also be related to the binding energy of the hydrogen molecule (E) and the temperature (T) using the following equation:

Kd = e^(-E/kT)

Where k is the Boltzmann constant.

By combining the above two equations:

[H]^2 / [H2] = e^(-E/kT)

And since we know that [H] = [H2]/2

[H2]/4 = e^(-E/kT)

Now we can solve for T:

T = -E/k * ln([H2]/4)

Given that the binding energy of hydrogen molecule is 4.48 eV and k = 8.62 x 10^-5 eV/K.

T = 4.48 / 8.62 x 10^-5 * ln([H2]/4)

The pressure is not used in the above calculation. The temperature is related to the binding energy and the dissociation equilibrium constant which is dependent on the concentration of the hydrogen molecules and atoms.

It is important to note that the above calculation is a rough estimate and the temperature of a stellar atmosphere can be affected by many other factors such as radiation, convection, and other chemical reactions.

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He travels a distance of 300 m in that time interval.

We know that we can be able to obtain the time that he travelled if we use the formula that applies to the average speed. In this case we are told in  a time of 30 seconds, he increases his speed steadily from 8.3 m/s to 11.7 m/s.

Thus;

distance = Average speed/Time

Distance = (8.3 + 11.7/2) * 30

Distance = 300 m

We can see that he would move through a distance of well about 300m as the velocity is increasing as shown here.

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a) The gravitational force on the earth is 81 times that on the moon

b) The slope of the graph is the universal gravitational constant

c) The force is  5 * 10^19 N

d) The acceleration due to gravity is the same.

We know that the term gravity has to do with the force that acts on any object that we can be able to find on the surface of the earth. We know that the larger the mass of the object, the greater the magnitude of the gravitational force on the object.

We can see that in the question the distance between the earth and the moon was halved hence the force would now be one quarter of its initial value hence;

New force = 1/4 * 2 * 10^20

= 5 * 10^19 N

The acceleration  due to gravity does not change hence it is the same both at the surface and near the surface.

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Imagine the same bouncing basketball as it is moving down, it is having a negative acceleration, as it is moving down.

This is because acceleration is defined as the rate of change of velocity, and negative acceleration occurs when an object's velocity is decreasing. The basketball is moving down, it is losing height, and its velocity is decreasing, thus it has a negative acceleration. To be more precise, the negative acceleration of the basketball is due to the force of gravity acting upon it, which is pulling it down toward the earth. The force of gravity is a constant force, acting downward, and it causes all objects to accelerate downward at a rate of 9.8 m/s^2 (on the surface of the earth). ​

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The potential difference between the plates is 2156V Plate B has the higher potential because the electron moves from A to B as it acquire its own kinetic energy.

Given kinetic energy of electron (KE) = 3.45×10^−16 J

The electron is accelerated from plate A to plate B by an electric field.

We know kinetic energy = charge of electron x potential difference.

So, KE = -qV where q = 1.6 x 10^-19C which is charge of electron.

-3.45×10^−16 = -1.6 x 10^-19 x V

V = 3.45×10^−16/1.6 x 10^-19 =  2156V

Sine the potential difference is positive we can say that plate B has higher potential that plate A.

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The charge on the sphere is found to be 1.5 x 10⁻¹⁵ C.

The proton is orbiting the surface of the charged sphere of radius 4.43 cm and the speed of the proton is 3.22 x 10⁵m/s. The charge on proton is 1.6 x 10⁻¹⁹ C and the mass of the proton is 1.67 x 10⁻²⁷ kg.

Now,

For revolving,

Electrostatic forces = Centripetal force

KQq/r² = Mv²/r

Here,

R is the radius of the orbit,

Q and q are the charges of the sphere and the proton,

M is the mass of the proton.

Putting values,

9 x 10⁹ x Q x 1.6 x 10⁻¹⁹/0.0443 = 1.67 x 10⁻²⁷ x 3.22 x 10⁵

Q = 1.5 x 10⁻¹⁵ C

So, the charge on the sphere is 1.5 x 10⁻¹⁵ C.

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Equations of motion for constant acceleration to find maximum elevation, time and velocity are as follows :

y = yi + vi*t + (1/2)*a*t^2

vf = vi + a*t

V^2= U^2 + 2 a s

Givens:

vf = 0 m/s (the ball stops at the time in question)

yi = 0 m (or assume whatever initial height you wish, and add to the final answer)

a = -9.8 m/s (we can add more significant digits, but this is close enough)

b) The maximum elevation above the roof the ball will reach can be found by,  y= yi + vi*t + (1/2)*a*t^2

c)Time to reach maximum height can be obtained from v = u + at

     at = v - u

therefore, time required is t = v - u / a

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The ability to do activities for more than a few minutes is called aerobic ability.

Endurance is the ability to maintain an activity for extended periods of time and usually refers to aerobic ability. For example, children can play actively for hours. We need this ability to perform repetitive activities of daily living, such as stirring food while cooking, using a hair dryer to dry our hair, or walking up steps.

Recreational and job-related activity also often require a high level of endurance. Local muscle endurance is best defined as the ability to resist muscular fatigue and demonstrate how a given type of contraction can be sustained, typically measured in terms of the number of repetitions.

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The radius of a planet is 7.01 times of earth's. The volume of the planet is 344.47 times bigger than the volume of earth.

Let's assume that the shape of both planets are perfect spheres. The volume of a sphere is given by:

V = 4/3 πr³

Where r is the radius.

Hence, the volume is directly proportional to r³, or:

V ∝ r³

Let:

r_p = radius of the planet

r_e = radius of earth

Then,

r_p = 7.01 r_e

Using direct proportionality:

V_p : V_e = r³_p :  r³_e

V_p : V_e = (7.01 r_e)³ :  r³_e

V_p : V_e = 344.47 : 1

V_p = 344.47 V_e

Therefore, the volume of the planet is 344.47 times bigger than earth's.

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The work done by the man in lifting his body is 493.92J.

What is work done?

When an object is moved over a specific distance by an external force, the quantity of energy transferred to it is called work. The quantity of energy transmitted to an object through work is referred to as the work done on the thing. Since the man raises his body a vertical distance of 0.63m to go from stage 1 to stage 2 he is doing work against gravity. So, the work done in this case is given by W=mgh where m is the mass of the body, g is the acceleration due to gravity and h is the height raised by the body.

Now, according to the question

Mass of the man=80Kg

Height raised by the man=0.63m

Acceleration due to gravity=9.8

On substituting the values in the given formula, we get the work done as

W=80×9.8×0.63

W=493.92J

Therefore, the work done by the man is 493.92J.

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To control a camera on the moon to perfectly follow the lunar lander as it descended from the surface, engineers used a combination of tracking systems and software.

First, they used a tracking camera to monitor the descent of the lander. This camera was mounted in a fixed position, and was able to track the movement of the lander in real-time.

Additionally, engineers used software to calculate the exact trajectory of the lander and to calculate the corresponding camera movements. This allowed the camera to perfectly follow the lunar lander as it descended from the surface.

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Therefore, the highest fluid index of refraction at which we can still produce entire internal reflection is 1.34.

A wave bends when it reaches a medium with a different speed, which is known as refraction. When light travels through a slow medium after traveling through a fast medium, the light ray is bent in the direction of the boundary between the two media.

When light travels from one material to another and changes direction or bends, this is known as refraction. When light strikes water, for instance, it refracts after passing through the air. This may cause a straw in a glass of water to seem twisted at the surface.

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The net force applied by to the box is 6264.2 N, if the mass of the box is 50.0 kg and it comes to rest in 0.05 second.

Free fall height of the box, h = 2.0 m

Free fall velocity of the box just before touching the ground, v = √(2gh)

= √(2×9.81×2) = 6.264 m/s

Final velocity or the box after 0.05 sec, = 0

Time to stop, t = 0.05 sec

We know that the applied force is equal to the rate of change of momentum.

F = m(v-u)/t

F = 50 × (6.264-0)/0.05 = 6264.2

So, F = 6264.2 N

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At the height of 3h/4 meter, the velocity of the baseball will be half of its original velocity v, if it reaches a maximum height h.

Maximum height reached by the baseball, = h

The velocity of the ball at maximum height v₁ = 0

Initial velocity, = v

by the 3rd equation of motion, v₁² = v² - 2gh

0 = v² - 2gh

v² = 2gh

Now let the height at which the velocity will be half of the original velocity = h₁

Then by the 3rd equation of motion,

(v/2)² = v² - 2gh₁

2gh₁ = v² - v²/4 = 3v²/4

h₁ = 3v²/8g

We know v² = 2gh, So

h₁ = 3(2gh)/8g = 3h/4 m.

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The height the ball reaches when the boy throws the ball up in the air is 12.6m.

Given the maximum horizontal distance a boy can throw a ball (s) = 50m

The initial speed of the ball = um/s

The maximum height reached = ym

The motion in vertical direction is given as:

y = ut1 + 1/2at1^2 where v is the speed in vertical direction and a is the acceleration and t1 is the time taken to travel upward distance.

s = 0 + gt^2/2 then t = √2s/g = √2 x 50/9.8 = 3.18s

We know that as the velocity is same at any point in distance travelled distance = speed x time = 50m

So t1 = t/2 as it covers half time when reached maximum height.

The velocity (v) = gt = 9.8 x 3.18 = 31.164m/s

The maximum height reached (y) = vt1 +  1/2gt1^2 where t1 - t/2

y = vt/2 + 1/2g(t/2)^2

y = 50/2 + 1/2(9.8)(3.18/2)^2

y = 12.6m

Hence the maximum height the ball can be thrown is 12.6m.

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The work done by the car is equal to the change in kinetic energy.

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Answer:A

Explanation:

The magnitude of the charge with a smaller magnitude is 2.42 × 10⁴ Coulomb. The result is obtained by using the formula for electrostatic force.

The formula for electrostatic force is

F = kq₁q₂/r²

Where

Two point charges are separated.

We have

Find  the smaller magnitude of the charge!

The smaller magnitude of the charge is q₂.

We use the formula above to find q₂.

F = kq₁q₂/r²

85 = (9×10⁹)2q₂q₂/(0.6)²

85 = (18×10⁹)q₂²/0.36

q₂² = (18×10⁹)/(85×0.36)

q₂² = (18×10⁹)/30.6

q₂² = 5.88 × 10⁸

q₂ = √(5.88 × 10⁸)

q₂ = 2.42 × 10⁴ Coulomb

Hence, the smaller magnitude of the charge is 2.42 × 10⁴ Coulomb.

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Answer:

The object's mass

The object's acceleration

Explanation:

To determine how much force was used on an object, we need to know the object's mass and acceleration. The force acting on an object can be calculated using the equation force = mass x acceleration (F = m*a) So, the more massive the object is, the more force is required to accelerate it, and the greater the acceleration is, the greater the force applied to the object.

A thermometer sensitive is a bulb with a thin glass wall. so, option (a) is correct.

What is thermometer?

A thermometer is a device that measures temperature. The three most used ways to measure temperature are in degrees Celsius, degrees Fahrenheit, and degrees kelvin. The metric system incorporates the Celsius scale.

What is thermometer sensitive?

The degree of change in a thermometer's thermometric property for a unit change in temperature is referred to as its sensitivity. It speaks of the smallest temperature difference that may be seen or quantified. The difference between the maximum and least temperature is referred to as a thermometer's range.

Therefore, a thermometer sensitive is a bulb with a thin glass wall. so, option (a) is correct.

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The new angular speed of the merry go round after the man came to the point 0m on the merry go round will be 1.37 rev/s.

The merry go round is rotating at the speed of 0.27 rev/s with a man of mass 85 kg on it at a pint of 2.2 m form he center of the axis of rotation.

We have to find the new speed of the merry go round when the person walk to the center of the circle.

The mass and radius of the merry go round are given to be 67 kg and 2.2m.

Using the conservation of momentum as the system is not under any external force,

Initial momentum = final momentum

I₁W₁ = I₂W₂

I₁ and W₁ are initial moment of inertia and angular speed, I₁ and W₁ are final moment of inertia and angular speed.

Putting values,

(85+67)(2.2)₂0.27 = (67)W₂

1.34 rev/s = W₂

So, the final angular speed is 1.34 rev/s.

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The beam scenario with a 7cm sector size and a scanning rate of 13cm will have better resolution.

This is because the more modest sector size considers a more engaged beam, and the faster scanning rate takes into account more information to be gathered in a more limited measure of time. This results in a higher-resolution image.

In the first scenario, the temporal resolution is 10cm sector size and scanning at 18cm. In this scenario, the sector size is 10cm, which means that the beam is 10cm wide.

The scanning rate is 18cm per second, which means that the beam scans 18cm in one second. The slower scanning rate in this scenario will result in a lower temporal resolution because less data are being collected in the same amount of time.

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The magnitude of resultant force of the two forces (40 N and 30 N) at right angles is 50 N.  

The Pythagorean theorem is used to calculate the consequence of two forces acting at right angles to one another. According to the Pythagorean theorem, the square of the hypotenuse, the longest side opposite the right angle, in a right triangle, equals the sum of the squares of the other two sides. (the two shorter sides).

In this instance, the two forces combine to form the triangle's two shorter sides, with the hypotenuse as the third force.

Therefore, the amount of the resulting force can be determined as follows:

⇒ √(40² + 30²) = √(1600 + 900) = √2500 = 50 N

Hence, the magnitude of the resultant force of the two forces (40 N and 30 N) at right angles is 50 N.

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