an artificial satellite is moving in a circular orbit of radius 36000 kilometre calculate its speed if it takes 24 hours to revolve around the earth ​

Answers

Answer 1

Explanation:

9420 km/hr is the correct answer

Hope this helps...☺

An Artificial Satellite Is Moving In A Circular Orbit Of Radius 36000 Kilometre Calculate Its Speed If

Related Questions

a 1600 kg car rounds a curve of radius 71 m banked at an angle of 15, What is the magnitude of the friction force required for the car to travel at 86 km/h

Answers

Answer:

The frictional force required for the car to travel is 8,365.01 N

Explanation:

Given;

mass of the car, m = 1600 kg

radius of the curved road, r = 71 m

banking angle, θ = 15⁰

velocity of the car, v = 86 km/h = 86/3.6 = 23.89 m/s

The two forces acting on the are:

1.  the parallel force to the banked plane

2. the centripetal force pushing the car up the banked plane

To keep the car traveling at 86 km/h;

frictional force + parallel force to the plane = centripetal force pushing the car up the banked plane

The parallel force to the banked plane:

F = mgsinθ

F = 1600 x 9.8 x sin(15⁰)

F = 4,057.98 N

The centripetal force pushing the car up the banked plane:

[tex]F_c= (\frac{mv^2}{r} )cos(\theta)\\\\F_c = (\frac{1600 \times 23.89^2}{71} )cos(15^0)\\\\F_c = 12,422.99 \ N[/tex]

The frictional force required for the car to travel:

[tex]F_k = F_c - F\\\\F_k = 12,422.99 \ N - 4,057.98 \ N\\\\F_k = 8,365.01 \ N[/tex]

Therefore, the frictional force required for the car to travel is 8,365.01 N

The power in an electrical circuit is given by the equation P= RR, where /is the current flowing through the circuit and Ris the resistance of the circuit. What is the current in a circuit that has a resistance of 100 ohms and a power of 15 watts?

[pleas ee helpppp)​

Answers

I= 0.39 A

OPTION B is the correct answer.

FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN

Answers

Clouds? I am not sure of your options!

From the water whole water cycle starts again.

Most possibly water should be the answer.

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?

A) 1.67 x 10^-4 s^-1

B) 5.43 x 10^-4 s^-1

C) 1.40 x 10^-4 s^-1

D) 2.22 x 10^-4 s^-1

Answers

OPTION C is the correct answer.

Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms​

Answers

Answer:

Sorry I don't know the answer

2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time​

Answers

Answer:

(c) at rest

Explanation:

Given

See attachment for the distance time graph

Required

What does the graph illustrate?

From the graph, we can see that the line of distance is a horizontal line.

This suggests that a time increases, the distance remains unchanged

When distance remains unchanged over time, then it means the object is at rest.

Hence, (c) is correct

The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ

Answers

Answer:

For (a): The chemical shift is [tex]2.08\delta[/tex]

For (b): The chemical shift is [tex]9.85\delta[/tex]

For (c): The chemical shift is [tex]7.5\delta[/tex]

Explanation:

To calculate the chemical shift, we use the equation:

[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]

Given value of spectrometer frequency = 200 MHz

For (a):

Given peak position = 416 Hz

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]

For (b):

Given peak position = [tex]1.97\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]

For (c):

Given peak position = [tex]1.50\times 10^3 Hz[/tex]

Putting values in above equation, we get:

[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]

how do you use the coefficient to calculate the number of atoms in each molecule?​

Answers

wait is there supposed to be a picture here?

Answer:

To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)

Explanation:

Calculating Acceleration
Initial
velocity
Time to travel
0.25 m
Final
velocity
Acceleration
Time to travel
0.50 m
# of
washers
11
(m/s)
V2
(m/s)
ti
(s)
t₂
(s)
a = (v2 - v4)/(t2-tı)
(m/s)
1
0.11
0.28
2.23
3.13
0.19
2
0.13
0.36
1.92
2.61
The acceleration of the car with two washers added to the string would be

Answers

I can not even read this question.

What are you trying to even say?

The acceleration of the car with two (2) washers added is equal to 0.33 [tex]m/s^2[/tex].

Given the following data:

Initial velocity = 0.13 m/s.Final velocity = 0.36 m/s.Initial time = 1.92 seconds.Final time = 2.61 seconds.

What is an acceleration?

An acceleration can be defined as the rate of change of velocity of an object with respect to time and it is measured in meter per seconds square.

How to calculate average acceleration.

In Science, the average acceleration of an object is calculated by subtracting its initial velocity from the final velocity and dividing by the change in time for the given interval.

Mathematically, average acceleration is given by this formula:

[tex]a = \frac{V\;-\;U}{t_f-t_i}[/tex]

Where:  

V is the final velocity.U is the initial velocity.[tex]t_i[/tex]initial time measured in seconds.[tex]t_f[/tex] final time measured in seconds.

Substituting the given parameters into the formula, we have;

[tex]a = \frac{0.36\;-\;0.13}{2.61\;-\;1.92}\\\\a=\frac{0.23}{0.69}[/tex]

a = 0.33 [tex]m/s^2[/tex]

Read more on acceleration here: brainly.com/question/24728358

g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?

Answers

Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field

Explanation:

While interpreting the data in NMR, the positions of signals are studied.

The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.

The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.

So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]

Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons

So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]

An electric device, which heats water by immersing a resistance wire in the water, generates 20 cal of heat

per second when an electric potential difference of 6 V is placed across its leads. What is the resistance in Ω

of the heater wire? (Note: 1 cal = 4.186 J)

Select one:
a. 0.86
b. 0.17
c. 0.29
d. 0.43

Answers

Answer:

1 cal/s =4.184w

p=50 cal/s =2093w

v=12v

P = V*I

I =P/V

I = 17.43 A

P =1²*R

R = P/I²

R = 0.68

Lighting is the movement of?

Answers

Explanation:

Movement:refers to the changing in the lights whether it be a change in intensity, color or direction of origin.

Which two statements are true for reversible reactions that reach dynamic
equilibrium?
I A. The products of the forward and backward reactions remain
constant at equilibrium.
B. The products of the forward reaction form more quickly than its
reactants.
C. The rate of the forward reaction is greater than the rate of the
backward reaction.
- D. The rate of the forward reaction is equal to the rate of the
backward reaction at equilibrium.

Answers

Answer:

Explanation:

In a reversible reaction which has reached dynamic equilibrium , rate of forward reaction is equal to rate of backward reaction .

Following is a reversible chemical reaction .

A + B = C + D

Rate of forward reaction = k₁ x [ A ] x [ B ]

Rate of backward reaction = k₂ x [ C ]  x [ D ]

k₁ x [ A ] x [ B ] =  k₂ x [ C ]  x [ D ]

[ A ] x [ B ] = k₂ / k₁  [ C ]  x [ D ]

[ A ] x [ B ] = k   [ C ]  x [ D ]

The products of the forward and backward reactions remain

constant at equilibrium.

Hence option A and D are correct statement .

A laser emits a single 3.0-ms pulse of light that has a frequency of 2.83E11 Hz and a total power of 65000 W. How many photons are in the pulse? Please provide all equations and work.

6.0E23
1.0E24
2.4E25
3.6E25
4.8E26

Answers

Answer:

The number of photons in the pulse is 1.04 x 10²⁴

Explanation:

Given;

frequency of the emitted photons, f = 2.83 x 10¹¹ Hz

duration of the incident light, t = 3 ms = 3 x 10⁻³ s

power of the incident light, P = 65,000 W

The energy of each photon emitted is calculated as;

E = hf

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ Js

E =  6.626 x 10⁻³⁴ x  2.83 x 10¹¹

E = 1.875 x 10⁻²² J

let the number of photons in the pulse = n

n(E)= Power x time

[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]

pls help! George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. What is the
power applied to push this wheelbarrow?
A. 1.2 watts
B. 3.4 watts
C. 6.9 watts
D. 13 watts

Answers

Answer:

C. 6.9 watts

Explanation:

Power = work/time

if work = force×distance...

Then... power= (force×distance)/time

Power = (20×12)/35

= 6.9 watts

The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?

Answers

Answer: The energy released as thermal energy is 6.5 J

Explanation:

Energy stored by the spider when it relaxes is given by:

[tex]E_o=\text{Resilience}\times \text{Work}[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\text{Work done}-E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5J[/tex]

Hence, the energy released as thermal energy is 6.5 J

The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J

What is thermal energy?

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

Energy stored by the spider when it relaxes is given by:

[tex]\rm E_o=Resilience \ \times Work[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]\rm E_o=0.35\times 10[/tex]

[tex]E_o=3.5\ J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\rm Work done -E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5\ J[/tex]

Hence, the energy released as thermal energy is 6.5 J

To know more about thermal energy follow

https://brainly.com/question/19666326

Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface

Answers

Answer:

Shiny surface.

Explanation:

We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.

Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.

This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,

A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.

Answers

Answer:

Explanation:

From the given information:

The initial PE [tex](PE)_i[/tex] = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = [tex]P.E_f -P.E_i[/tex]

ΔP.E = 0 - [tex](PE)_i[/tex]

ΔP.E = [tex]-P.E_i[/tex]

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]

[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]

this can be re-written as:

[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]

[tex]\Delta U =(0.70) (490.5)[/tex]

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s

Answers

Answer:

[tex]\boxed {\boxed {\sf 40 \ kilograms}}[/tex]

Explanation:

Momentum is the product of velocity and mass. The formula is:

[tex]p=m*v[/tex]

We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.

[tex]200 \ kg \ m/s = m * 5.0 \ m/s[/tex]

We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.

[tex]\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}[/tex]

[tex]\frac{200 \ kg \ m/s}{5.0 \ m/s}=m[/tex]

The units of meters per second (m/s) cancel.

[tex]\frac{200 \ kg}{5.0 } =m[/tex]

[tex]40 \ kg = m[/tex]

The falling rock has a mass of 40 kilograms.

A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.

Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?

Answers

Answer:

a)   μ = 0.0136, b)   F = 22.8 N

Explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

        p₀ = m v₀

final instant. After inelastic shock

        p_f = (m + M) v

the moment is preserved

        p₀ = p_f

        m v₀ = (m + M) v

        v = [tex]\frac{m}{m + M} \ v_o[/tex]

We look for the speed of the block with the bullet inside

        v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]

        v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

         W = ΔK

          W = ½ (m + M) [tex]v_f^2[/tex]  - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

         W = - fr x

we substitute

           - fr x = 0 - ½ (m + M) vo²

           fr = ½ (m + M) v₀² / x

         

the friction force is

          fr = μ N

          μ = fr / N

equilibrium condition

          N - W = 0

          N = W

          N = (m + M) g

we substitute

         μ = ½ v₀² / x g

we calculate

          μ = ½ 0.448 ^ 2 / 0.75 9.8

          μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

          W = ΔK

initial block velocity is zero vo = 0

         F x₁ = ½ M v² - 0

         F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]

         F = ½ 2.50 0.448² / 0.0110

         F = 22.8 N

please help very easy 5th grade work giving brainliest

Answers

Answer:

the answer is option B because opposit sides of the magnets attract each other

An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth

Answers

Answer:

0 N

Explanation:

This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N

If you exert a force of 5 N into a nutcracker, and it outputs a force of 20 N, what is the mechanical advantage of the nutcracker. Show formula PLSSS HELPPPP!!! i'll make you brainliest

Answers

Answer: 4

Explanation:

MA = output force / input force

MA = 20 / 5

MA = 4

Hope this helps.  Please mark brainliest.

PLS HELP ME 100 POINTS PLS I NEED HELP QUICK PLS


For this project, you are expected to submit the following:
1. Your Student Guide with completed Student Worksheet
2. Your scale model of the solar system
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Conduct research on the actual sizes of the planets.
a) Do research to find the actual sizes of the Sun and the planets. This information is typically represented as diameter in kilometers (km). Recall that diameter is the length of the imaginary straight line from one side of a figure, such as a sphere, to the opposite side of the figure. This line passes through the center of the figure.
b) Record the actual diameters of the Sun and the planets in the first column of the table in the Student Worksheet.
c) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 3: Determine the scaled sizes of the planets.
a) Go to a reliable website to find a solar system model calculator.
b) Decide how big you want the Sun in your model to be. For example, you could assign your Sun to be 300 mm. Input this figure in the calculator, and the calculator will determine the diameters of the eight planets for you. You want to make sure that the Sun is big enough so that the smallest planet will still be big enough to draw.
c) Record information from the calculator in the second column of the table in the Student Worksheet.
d) Copy the link of the website you used into the space provided in the Student Worksheet.
Step 4: Create a scale model of the solar system.
a) Draw and cut construction paper models of the Sun and the planets using the scaled measurements from the table.
b) Glue the models on the poster board. You can glue or tape poster boards together if necessary. Be sure to put the Sun in the center and to put the planets and a drawing of their orbits in order from nearest to farthest from the Sun.
Note: Remember that in this model, the diameter of the planets is scaled but the distance of the planets from the Sun is not. That means your model does not accurately represent the distances of the planets from the Sun so you need not worry about these measurements.
c) Label the Sun and the planets.
d) Put an attention-catching title above or below your model.
e) Write your name on the back of your poster board.
Step 5: Complete the Student Worksheet.
a) Make sure the table in the Student Worksheet is complete.
b) Answer the questions in the Student Worksheet.
c) Check to make sure you added the sources you used for this project in the Student Worksheet.
Step 6: Evaluate your project using this checklist.
If you can check each of the following boxes, you are ready to submit your project.
 Did you conduct research to find the actual size of the Sun and the planets? Did you record this information in the table in the Student Worksheet?
 Did you use a solar system model calculator to determine the scaled size of the Sun and planets? Did you record this information in the Student Worksheet?
 Did you add the links of the websites you used for this project to the Student Worksheet?
 Did you use the scaled sizes to create models of the Sun and the planets?
 Did you put your model together in a way that represents the solar system (Sun in the center and planets in order from nearest to farthest from the Sun)?
 Did you label each component of your model?
 Did you add an attention-catching title above or below your model?
 Did you write your name on the back of your poster board?
 Did you complete the Student Worksheet at the end of this guide?
Step 7: Revise and submit your project.
a) If you were unable to check off all the requirements on the checklist, go back and make sure that your project is complete. Save your project before submitting it.
b) Turn in your scale model of the solar system to your teacher. Be sure that your name is on it.
c) Submit your Student Guide through the virtual classroom.
d) Congratulations! You have completed your project.

Answers

Answer

I hope this help....

Explanation:

Answer:

Hope this helps

Explanation:

calculate the correct fuse that should be used for a 230V,1KW electric hair dryer.​

Answers

Answer: 4 A

Explanation:

Given

Voltage [tex]V=230\ V[/tex]

Power [tex]P=1\ kW[/tex]

Power is given by [tex]P=VI\\[/tex]

Insert the values

[tex]\Rightarrow 1000=230\times I\\\\\Rightarrow I=\dfrac{1000}{230}\\\\\Rightarrow I=3.84\ A[/tex]

The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.

A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV

Answers

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Answers

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]

[tex]v_1t= 0.35[/tex]  ...........(i)

[tex]v_0t= 1.09[/tex]  ...........(ii)

Dividing the equation (ii) by (i)

[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]

∴ [tex]v_0=3.11 \ v_1[/tex]

Now loss of energy  = change in the kinetic energy

[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]

[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]

[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]

If f is average friction force, then

(f)(L) = W

(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

(f)  = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

The Average force of friction is ( F )  = 7.307 * 10⁻³ v₀²

Given data:

Predicted range ( v₁t ) = 0.3503 m

Actual range ( v₀t ) = 1.09 m

mass = 16.3 g

First step : Determine the value of  V₀

[tex]t = \sqrt{\frac{2H}{g} }[/tex]    ,    v₁t  =  0.3503 ,    ( v₀t ) = 1.09 m

To obtain the value of  V₀  

Divide ( v₀t ) by ( v₁t )  =  1.09 / 0.3503 = 3.11 v₁

V₀ = 3.11 v₁

Next step : Determine the average force of friction ( f )

given that loss of energy results in a change in kinetic energy

W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]

    = 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]

W = 7.307 * 10⁻³ v₀²

Average force of friction = W / Actual length

                                         = 7.307 * 10⁻³ v₀² / 1  

∴ Average force of friction ( F )  = 7.307 * 10⁻³ v₀²

Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²

Learn more about average force of friction : https://brainly.com/question/16207943

Your question has some missing data below are the missing data related to your question

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

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Answers

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66,7 seconds

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the formula for height/distance is : S=v.t

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Answer:

564

Explanation:

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A. Nuclear energy
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SUBMI

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Answer:

c

Explanation:

in food and batteries chemical energy is stored :) hope this helped

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