does net force stay the same when a massless pulley is replaced by a pulley with mass

Does Net Force Stay The Same When A Massless Pulley Is Replaced By A Pulley With Mass

Answers

Answer 1
It pulls gravity so no

Related Questions

The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.

Answers

Answer: 1 cal is 4.186 J, 1 kcal = 4186 J   A : 1014 m , B  200 m

Explanation:   A) Work done by climber is change in potential energy.

W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.

Solve h  =  160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m

B  Energy is only 20 %  :   Then  h  =  0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.

Actually, muscles also produce heat from most of the energy provided by food.

Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.

Answers

Answer:

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

Explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]

Making K the subject;

[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]

[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]

[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
the acceleration due to gravity is 9.8 m/s2, what is the crate's potential energy
at this point?

Answers

Answer:

[tex]\boxed {\boxed {\sf 29,400 \ Joules}}[/tex]

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

[tex]E_P= m \times g \times h[/tex]

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg g= 9.8 m/s²h= 20 m

Substitute the values into the formula.

[tex]E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m[/tex]

Multiply the three numbers and their units together.

[tex]E_p=1470 \ kg*m/s^2 \times 20 m[/tex]

[tex]E_p=29400 \ kg*m^2/s^2[/tex]

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

[tex]E_p= 29,400 \ J[/tex]

The crate has 29,400 Joules of potential energy.

Answer:

29,400 J

Explanation:

did the quiz <3

A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?

Answers

Answer:

3.73 * 10^16   photons/sec

Explanation:

power supply = 3.0 V

Emits 440 nm blue light

current in LED = 11 mA

efficiency of LED = 51%

Calculate the number of photons per second the LED will emit

first step : calculate the energy of the Photon

E = hc / λ

   =(  6.62 * 10^-34 * 3 * 10^8 )  / 440 * 10^-9

   = 0.0451 * 10^-17  J

Next :

Number of Photon =( power supply * efficiency * current ) / energy of photon

= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17

= 3.73 * 10^16 photons/sec

Suppose that 2 J of work is needed to stretch a spring from its natural length of 28 cm to a length of 43 cm. (a) How much work is needed to stretch the spring from 33 cm to 35 cm

Answers

Answer:

0.035 J

Explanation:

Applying,

W = ke²/2.............. Equation 1

Where W = workdone by the stretching the spring, k = spring constant, e = extension.

make k the subject of the equation

k = 2W/e²............... Equation 2

From the question

Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m

Substitute these values into equation 2

k = (2×2)/(0.15²)

k = 177.78 N/m

Hence, work need to stretch the spring from 33 cm to 35 cm

therefore,

e = 35-33 = 2 cm = 0.02 m

Substitute into equation 1

W = 177.78(0.02²)/2

W = 0.035 J

I’ve been stuck please help !!

Answers

Answer:

The slope of the position time graph gives the velocity.

Explanation:

The slope of the position time graph gives the value of velocity.

In first graph,

The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts.  and more than the second part, so the initial velocity is more than the final velocity.

In the second graph,

The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.

A rectangular field is of length 42 cm and breadth 25 m. Find the area of the field in SI unit. EXPLAIN STEP BY STEP​

Answers

Answer:

the area of the rectangular field is 10.5 m²

Explanation:

Given;

length of the rectangular field, L = 42 cm = 0.42 m

breadth of the rectangular field, b = 25 m

The area of the rectangular field is calculated as follows;

Area = Length x breadth

Area = 0.42 m x 25 m

Area = 10.5 m²

Therefore, the area of the rectangular field is 10.5 m²

The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.​

Answers

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

What are 3 artificial and 2 natural sources of electromagnetic radiation?

Answers

Answer: its b bro

Explanation:

ajafa'jfbA'FJ

A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes 0.9 s for the wave to travel the 26 m to the opposite tower. If one meter of the rope has a mass of 0.28 kg, find the tension in the tightrope.

Answers

Answer:

the tension in the tightrope is 233.68 N

Explanation:

Given the data in the question;

Time taken to reach the opposite tower t = 0.9 s

Distance between the two towers S = 26 m

mass per one meter length =  0.28 kg

First we calculate the velocity;

Velocity V = Distance / time

we substitute

Velocity V = 26 m / 0.9 s

Velocity V = 28.889 m/s

We know that Velocity V can also be expressed as;

V = √( T / m )

we make T the subject of formula

V² = T / m

T = mV²

we substitute

T = 0.28 × ( 28.889 )²

T = 233.68 N

Therefore,  the tension in the tightrope is 233.68 N

Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to the difference of the temperature of the medium and the temperature of the object. Suppose a metal bar, initially at temperature 50 degrees Celsius, is placed in a room which is held at the constant temperature of 40 degrees Celsius. One minute later the bar has cooled to 40.18316 degrees . Write the differential equation that models the temperature in the bar (in degrees Celsius) as a function of time (in minutes). Hint: You will need to find the constant of proportionality. Start by calling the constant k and solving the initial value problem to obtain the temperature as a function of k and t . Then use the observed temperature after one minute to solve for k .

Answers

Answer:

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),

Explanation:

45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.

Answers

D okokokokokokok I’m right promise

Find the weight of a man whose mass is 40 kg on earth.


(also
write complete data plus proper formula).



Answers

I am sure it help you with that much ☺️

Explanation:

pleasae give me some thanks please good morning sister

From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.

Answers

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.

Answers

Answer:

a. 50km/hr.

b. 10km/hr

Explanation:

Average speed, which is calculated by dividing the total distance travelled by the time interval as follows:

Average speed = total distance travelled ÷ time

Average velocity is calculated by dividing the total displacement by the time interval as follows:

Average velocity = change in displacement (∆x) ÷ time (t)

According to this question, a motorist travels due North at 90 km/h for 2 hours. She then changes direction and travels West at 60 km/for 1 hour.

Total distance of this journey is 90 + 60 = 150

Total time taken = 1 + 2 = 3hours

Average speed = 150/3

= 50km/hr.

b.) Average velocity = x2 - x1/t

Average velocity = 90 - 60/3

= 30/3

= 10km/hr

Calculate the heat energy conducted per hour through the side walls of a cylindrical steel
boiler of 1.00 m diameter and 3.0 m long if the internal and external temperatures of the
walls are 140 °C and 40 °C respectively and the thickness of the walls is 6.0 mm. (Thermal
conductivity of steel, k = 42 Wm-4°C-4)

Answers

Explanation:

heat caoacity and heat is difference

The heat energy conducted per hour through the side walls of the cylindrical steel boiler  is 27708847 kJ.

What is thermal conductivity?

The rate at which heat is transported by conduction through a material's unit cross-section area when a temperature gradient exits perpendicular to the area is known as thermal conductivity.

In the International System of Units (SI), thermal conductivity is measured by Wm⁻¹K⁻¹.

Diameter of the cylindrical steel boiler: d = 1.00m.

Length  of the cylindrical steel boiler: l = 3.00m.

thickness of the walls is = 6.0 mm = 0.006 m

Temperature gradient is = (140-40) °C/0.006 m = 1666.67 °C/m

Thermal conductivity of steel,  = 42 W/m-°C.

Hence, the heat energy conducted per hour through the side walls of the cylindrical steel boiler = 42×3600×1666.67 ×2π×0.5(0.5+3.0) Joule

= 27708847 kJ

Learn more about  thermal conductivity here:

https://brainly.com/question/23897839

#SPJ2

An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?

Answers

Answer:

The angle is 4.1 rad.

         

Explanation:

The centripetal acceleration (α) is given by:

[tex] \alpha = \omega^{2} r [/tex]    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

[tex] a = \alpha r [/tex]      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

[tex] \omega^{2} r = 8.2\alpha r   [/tex]

[tex] \omega^{2} = 8.2\alpha [/tex]    (3)      

Now, we can find the angle with the following equation:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]

Where:

[tex] \omega_{f}[/tex]: is the final angular velocity                                                                              [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)

[tex]\Delta \theta[/tex]: is the angle

[tex] \omega^{2} = 2\alpha \Delta \theta [/tex]     (4)    

By entering equation (3) into (4) we can calculate the angle:

[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]

[tex] \Delta \theta = 4.1 rad [/tex]

Therefore, the angle is 4.1 rad.

I hope it helps you!                  

1. Lifting an elevator 18m takes 100kJ. If doing so takes 20s, what is the average power of the elevator during the process?

2. How much work can a 0.4 hp electric mixer do in 15 s?​

Answers

Answer:

1. Power = 5000 Watts

2. Workdone = 11185.5 Joules

Explanation:

Given the following data;

1. Distance = 18 m

Energy = 100 KJ = 100,000 Joules

Time = 20 seconds

To find the average power of the elevator;

Power = energy/time

Power = 100000/20

Power = 5000 Watts

2. Power = 0.4 HP

Time = 15 seconds

Conversion:

1 horsepower = 745.7 Watts

0.4 horsepower = 0.4 * 745.7 = 298.28 Watts

To find the amount of work done by the electric mixer;

Work done = power * time

Workdone = 745.7 * 15

Workdone = 11185.5 Joules

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 25 s interval

Answers

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.​

Answers

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

A professional quarterback throws a 0.40 kg football. what is the force of weight?

Answers

Answer:

3.92N

Explanation:

Force= mass×accelerarion due gravity

But mass= 0.40kg

acceleration due to gravity = 9.8 m/s^2

Force = 0.40×9.8

Force=3.92N

An unstretched ideal spring hangs vertically from a fixed support. A 0.4 kg object is then attached to the lower end of the spring. The object is pulled down to a distance of 0.35 m below the unstretched position and released from rest at time t= 0. A graph of the subsequent vertical position y of the lower end of the spring as a function of t is given above, where y= 0 when the spring was initially unstretched. At which time is the upward velocity of the object the greatest?

Answers

Answer:

The correct answer will be "0.25 sec".

Explanation:

The graph of the given question is attached below.

According to the graph of the question,

Time,

T = 1 sec

For the upward velocity,

⇒ [tex]t = \frac{T}{4}[/tex]

By putting the value, we get

⇒    [tex]=\frac{1}{4}[/tex]

⇒    [tex]=0.25 \ sec[/tex]

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

     

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

           

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced

Answers

The current is induced

At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height of 4.35 m above where it started. Using conservation of energy, find the height of the ball when it has a speed of 2.5 m/s.

Answers

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

How can i prove the conservation of mechanical energy?​

Answers

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

A hockey puck is sliding across the ice with an initial velocity of 25 m/s. If the coefficient of friction between the hockey puck and the ice is 0.08, how much time (in seconds) will it take before the hockey puck slides to a stop

Answers

Answer: 31.89seconds

Explanation:

Based on the information given, we are meant to calculate deceleration which will be:

t = V/a

where, a = mg

Therefore, t = V/mg

t = 25/0.08 × 9.8

t = 25/0.784

t = 31.89seconds

Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.

A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting

Answers

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons

Answers

Answer:

No of proton is 13 and nucleus is 13

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of 1.1 m/s, while Ender recoils with a velocity of -4.3 m/s. Determine the ratio Bonzo/mEnder of the masses of these two enemies.

Answers

Answer:

the ratio Bonzo/mEnder of the masses of these two enemies is 3.91

Explanation:

Given the data in the question;

Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s

Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s

the ratio Bonzo/mEnder of the masses of these two enemies = ?

Now, using the law of conservation of momentum.

momentum of both Bonzo and Ender are conserved

so

Initial momentum = final momentum

we have

0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]

[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]  ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]

we substitute

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ -4.3 m/s / 1.1 m/s ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ -3.9090 ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = 3.91

Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91

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