what is the speed of a boy moving around a circular park of radius 6m if he goes around the park in 20s

Answers

Answer 1

Answer:

1.884 meters per second

Explanation:

s = d/t

d = C = 2πr = 2π(6) = 37.68 m

t = 20 s

s = (37.68 m)/(20 s)

s = 1.884 m/s

Answer 2

The speed of the boy is 1.88 m/s while he is moving around a circular park of radius 6m and taking time 20 seconds.

What is speed?

Speed can be defined as a measurement of how quickly the distance an object traveled changes with time. Speed is a scalar quantity as it has magnitude but no direction as a unit of measurement.

An object that moves quickly and with high speed, traveling a lot of ground in a lesser time.

Given, the radius of the circular park, r = 6m,

The time taken to cover one round of the park, t = 20 s

The total distance traveled by the boy is equal to the circumference of the circular park, so

Circumference of park, C = d = 2πr = 2× 3.14 × 6 =37.68 m

The speed of the boy, S = d/t

S = 37.68 /20

S = 1.88 m/s

Therefore, the speed of the boy is equal to 1.88 m/s.

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Related Questions


A 25.0kg boy is sliding on a frictionless frozen lake at 5.00m/s to the north when he is struck by a 1.00kg
snowball moving at 15.0m/s from the west. If the snowball sticks to him, how fast, and in what direction,
does the boy move after the collision?

Answers

The final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east

The given parameters;

Mass of the boy, m₁ = 25 kgSpeed of the boy, u₁ = 5 m/sMass of the snowball, m₂ = 1.0 kgSpeed of the snow ball, u₂ = 15 m/s

The initial momentum of the boy is calculated as follows;

[tex]P_y = m_1 u_1\\\\P_y = 25 \times 5\\\\P_y = 125 \ kgm/s \ \ north[/tex]

The initial momentum of the snowball is calculated as follows;

[tex]P_x = m_2 u_2\\\\P_x = 1 \times 15 \ \\\\P_x = 15 \ kgm/s \ \ west[/tex]

The resultant momentum of the boy and the snowball after collision is calculate as follows;

[tex]P_f = \sqrt{P_y^2 + P_x^2} \\\\P_f = \sqrt{125^2 + 15^2} \\\\P_f = 125 .9 \ kgm/s[/tex]

The final velocity of the system boy-snowball system is calculated as;

[tex]v_f(m_1 + m_2)= P_f\\\\v_f = \frac{P_f}{m_1 + m_2} \\\\v_f = \frac{125.9}{25 + 1} \\\\v_f = 4.84 \ m/s[/tex]

The direction of the boy after the collision is calculated as follows;

[tex]\theta = tan^{-1}(\frac{v_y}{v_x} )\\\\\theta = tan^{-1} (\frac{5}{15} )\\\\\theta = 18.4 \ ^0[/tex]

Thus, the final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east

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A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s

Answers

The speed of the cannon after firing is 1.2 m/s

This can be solved using the law of conservation of momentum.

From the law of conservation of momentum,

⇒ The momentum of the projectile is equal to the momentum of the cannon.

MV = P................ Equation 1

⇒ Where :

M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.

⇒ make V the subject of the equation

V = P/M.................. Equation 2

From the question,

⇒ Given:

P = 600 kgm/sM = 500 kg

⇒ Substitute these values into equation 2

V = 600/500V = 1.2 m/s

Hence, The speed of the cannon after firing is 1.2 m/s

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9. P waves move faster than S waves A. True B. False

Answers

Answer: true
(I need 20 characters to submit the answer so I wrote this not to be confused with the answer)

Two forces act on a block as shown in the picture. What is the net force of the block?

30 N to the right

30 N to the left

10 N to the left 10

10 N to the right

Answers

Answer:

10 N to the left.

Explanation:

Since the forces are acting in opposite directions, you need to calculate the difference.

20 N - 10 N = 10 N

More force is being exerted to the left. Therefore, the net force is 10 N to the left.

Which of the following is NOT a function of the lens in the eye?
W
It can perform minor adjustments for distance.
It flattens when light rays from distant objects are to be focused.
It is a light receptor that generates nerve signals that are sent to the brain
It maintains its spherical shape to view nearby objects.

Answers

Answer:

the last one

Explanation:

The maximum speed with which an 1000 kg car makes a 180-degree turn is 10 m/s. The radius of the circle through which the car is turning is 24 m. Determine the force of friction and the coefficient of friction acting upon
the car.

Answers

109 is the answer There hope you do well

Need help asap

A scientist has invented a robot to work on the seabed. According to his calculation, the armour of the robot can withstand a maximum pressure of 10⁵ Pa exerted by the sea water. If the density of the sea water is 1025 kg/m3, what is the maximum depth of the seabed that this robot can work? [Given g = 9.81 m/s2 and rho water = 1000 kg/m3] ​

Answers

Answer:

Explanation:

Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?

Obviously the sea is salt water so we can ignore  ρ = 1000 kg/m³

1025 kg/m³(d  m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa

d = 9.9450535...

d = 10 meters

That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa

so the robot would be a hair above its pressure limit before it even got in the water.

Which statement is true for a series circuit

Answers

Answer: they have one path to flow

Explanation: share the same current

In a series circuit, the current is the same at each resistor. If the light bulbs are identical, then the resistance is the same for each resistor. The voltage drop (I•R) will be the same for each resistor since the current at and the resistance of each resistor is the same.

when the tides are especially weak it is called a _____ tide

Answers

Answer:

When the tides are especially weak, it is called a NEAP tide.

Explanation:

Hope this helps :)

Answer:

Neap

Explanation:

Answered!

1. A roller coaster with a mass of 800 kg sits stationary at the top of a section of track, 75 m above

the ground as shown. When the brake is released, it starts to roll down the track


2. For each height represented in the diagram, calculate the gravitational potential energy using

Ep = mgh. Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for

each of the heights of the roller coaster. (6 marks)


3. Assuming there is no friction, determine the mechanical kinetic energy using Ek = Etotal - Ep.

Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for each of

the heights of the roller coaster. (6 marks)


4. For each height represented in the diagram, calculate the velocity using = �2

. Show ONE

SAMPLE calculation in the calculations section below and fill in Table 1 for each of the heights of

the roller coaster. (6 marks)


5. Use your answers to graph how gravitational potential energy, mechanical kinetic energy, and

velocity change as the roller coaster changes height. Use different colours for the three lines on

the graph. Graph paper is provided below. (3 marks)


6. Repeat steps 1 – 5 above for a roller coaster cart that has a mass of 300 kg and enter your

results in Table 2.

Calculations:

800 kg roller coaster cart:

Sample calculation for gravitational potential energy:

Sample calculation for Mechanical kinetic energy:

Sample calculation for velocity:


300 kg roller coaster cart:

Sample calculation for gravitational potential energy:

Sample calculation for mechanical kinetic energy:

Sample calculation for velocity:


Results:

Table 1: Potential energy, kinetic energy, total energy, and velocity of the 800 kg roller coaster cart


Table 2: Potential energy, kinetic energy, total energy, velocity of the 300 kg roller coaster cart.


Graphs:

It’s graphing time. These graphs are a bit different than the ones you did in the

data analysis assignment at the beginning of the course. In this case you have

three things to graph on each graph. (One graph for the 800 kg roller coaster cart

and one graph for the 300 kg roller coaster cart.) You need to graph the

gravitational potential energy with respect to height, the mechanical kinetic

energy vs height, and the velocity vs height.

Let’s look at the energy graphs first. In this case both kinetic energy and

mechanical energy cover the same range of values. This means they can use the

same scale on the y-axis. So, you will use the left y-axis and the x-axis to graph

the kinetic energy vs height and the potential energy vs height. You will need a

legend to explain which line is which. Colour coding is a nice way to highlight this.

The velocity values are much different than the energy values. This means you

need a totally different scale. So, your left y-axis won’t work. You need to make a

second scale on the right y-axis for your velocity values. You will plot the points

the same way as normal, but you will use the numbers on the right-hand scale

instead. Again, be sure to add your velocity line to the legend with a separate

colour code.


Discussion Questions:

1. Describe the relationship between the gravitational potential energy and the mechanical kinetic

energy of the roller coaster on your graph. (2 marks)

2. Describe the shapes of each of the three lines in the graph. Explain why the velocity is different.

(4 marks)

3. Describe how mass affects the speed at the bottom of the roller coaster. (2 marks)

4. Describe how mass affects the gravitational potential energy and the mechanical kinetic energy

of the roller coaster. (2 marks)

5. At what point does the roller coaster have a maximum value for the following? Justify your

answer by explaining why. (2 marks each)

a. Gravitational potential energy

b. Mechanical energy

c. Velocity

6. In your calculations, you assumed that the roller coaster was frictionless. All real roller coasters

encounter friction. Describe how the actual values of the variables would differ, or not differ,

from your calculated values for a real roller coaster. (Hint: what form of energy would some of

the total energy be converted to if there was friction in the system?) (4 marks)

How you will be graded:

Grades will be based on answering questions to demonstrate an understanding of the material covered

in this unit. Point form answers are okay if ideas are complete and use vocabulary (Word Bank)

provided. For questions out of 4 marks, there are 4 responses expected.

Answers

Answer:

Give me some hint please

Based on the calculations, potential energy of this roller coaster at a height of 75 meters is equal to 588,000 Joules.

How to calculate potential energy?

Mathematically, potential energy is calculated by using this formula:

P.E = mgh

Where:

P.E represents potential energy.m is the mass.h is the height.g is acceleration due to gravity.

Note: Acceleration due to gravity is equal to 9.8 m/s².

At a height of 75 m, we have:

P.E = 800 × 9.8 × 75

P.E = 588,000 Joules.

At a height of 60 m, we have:

P.E = 800 × 9.8 × 60

P.E = 470,400 Joules.

At a height of 45 m, we have:

P.E = 800 × 9.8 × 45

P.E = 352,800 Joules.

At a height of 30 m, we have:

P.E = 800 × 9.8 × 30

P.E = 235,200 Joules.

At a height of 15 m, we have:

P.E = 800 × 9.8 × 15

P.E = 117,600 Joules.

In conclusion, we can deduce that the potential energy of this roller coaster decreases with a decrease in height.

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How many valence electrons do most stable atoms have?

A) one

B) two

C) four

D) eight

Answers

Answer:

D) 8

Explanation:

Due to the octet rule the most stable atoms will have 8 valence electrons.

An LED is useful because when a current passes through it, it gives out... what?
Enter your answer

Answers

An LED is useful because when a current passes through it, it gives out light.

An LED is useful because when a current passes through it, it gives out Light.

What is an LED?

LED, in full light-emitting diode, in electronics, a semiconductor device that emits infrared or visible light when charged with an electric current.

A light-emitting diode (LED) emits light by applying a forward current to the pn junction of a compound semiconductor.

When forward current is passed through the light-emitting diode, carriers (electrons and holes) move. The holes in the p-type region move to the n-type region and the electrons in the n-type region move to the p-type region. The injected carriers recombine, and the energy difference before and after recombination is released as light. The emitted light depends on the energy band gap (Eg) of the compound semiconductor.

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One indicator NOT related to domestic violence is
violent attitudes
good nutrition
violations of court orders
substance abuse

Answers

Answer:

good nutrition

Explanation:

I think that one's pretty self explanatory

True or False. What makes faith genuine is its object.

Answers

Answer:true

Explanation:

Why is it important that we learn about our solar system, the sun, and planets in our solar system other than Earth?

Answers

Answer:

To understand all the different kind of elements and how they work. Nature is a beautiful thing and it is important we understand how they survive and how we can help. The solar system is a mandatory nature. Learning about other planets also helps us feel safe and more knowledgable.

a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising

Answers

Answer:

when the ball is at its highest point

Explanation:

Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.

A woman skis from rest down a hill 20 m high. If friction is negligible, what is her speed at the bottom of the slope? Select one: O a. 20 m/s O b. 12 m/s O c. 400 m/s O d. 6 m/s​

Answers

Hi there!

We can use the work-energy theorem to solve.

Recall:

[tex]\large\boxed{E_i = E_f}}[/tex]

Initial energy = final energy

The initial energy is purely potential (she starts from rest), and, if we assign the bottom of the slope as the 0 line, her energy at the bottom is purely kinetic.

PE = mgh

KE = 1/2mv²

We can begin by setting the two equal:

mgh = 1/2mv²

Cancel out the mass and rearrange to solve for velocity:

2gh = v²

v = √2gh

Plug in given values and use g ≈ 10 m/s²:

v = √2(10)(20) = 20 m/s

What are the two types of force

Answers

Answer:

Forces can be divided into primarily into two types of forces:

Contact Forces. Non-contact Forces.

Explanation:

f. Protons and neutrons
2. TRUE or FALSE: An object that is positively charged contains all protons and no electrons
3. TRUE or FALSE: An object that is negatively charged could contain only electrons with me
accompanying protons
4. TRUE or FALSE: An object that is electrically neutral contains only neutrons.

Answers

An object which has more electron than proton is negatively charged, otherwise positively charged. Every statement is false

What is atom?

Atom is the smallest unit of the matter consist of the positive charged nucleus and the electrons which moves around it. The atom can not be divided further.

The atom of a matter is made by three elements-

1) Neutron-Neutron is the element of atom, which has zero charge.2) Proton-Proton is the element of atom, which has positive charge.3) Electron-Electron is the element of atom, which has negative charge.

For the Protons and neutrons, lets check all the statement wheather they are true or false.

2. An object that is positively charged contains all protons and no electrons- An object which is positively charged, has more number of proton than electron. This is a false statement.3. An object that is negatively charged could contain only electrons with me accompanying protons- An object, which is negatively charged, has more number of electron than proton. This is a false statement.4. An object that is electrically neutral contains only neutrons-The number of electron and proton is equal in an electrically neutral object. This is a false statement.

Thus, every statement is false as a object which has more electron than proton is negatively charged, otherwise positively charged.

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How is Compression Force Measured?

don't just copy theanswer please​

Answers

Compression force can be measured with a force gage or load cell. 

A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position �� as a function of time . What are the oscillation’s: a. period; b. frequency; c. angular frequency; d. amplitude; and e. What is the force constant of the spring?

Answers

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(c) The angular frequency of the oscillation is 7.885 rad/s.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

Mass of the ball, m = 2.4 kg

From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

The period of the oscillation is the time taken to make one complete cycle.

T = 0.8 s

The frequency of the oscillation is calculated as follows;

[tex]f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz[/tex]

The angular frequency of the oscillation is calculated as follows;

[tex]\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s[/tex]

The force constant of the spring is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m[/tex]

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A box of mass 7.7 - kg is accelerated from rest across a floor at a rate of 2.6 m/s2 for 18.5 s. Find the net work done on the box

Answers

Answer:

Explanation:

The net work will change the kinetic energy

W = ½mv² = ½m(at)² = ½ma²t²

W = ½(7.7)2.6²(18.5²) = 8907.3985 = 89 kJ

A textbook weighs 34 N on the surface of the Earth. What is the book’s mass on Earth’s surface?

Answers

Answer:

About 3.47kg

Explanation:

Recall that weight is equal to mass times acceleration.

In this case, our acceleration is due to gravity which on earth is about 9.8m/s/s

So we have 34N=9.8 *mass, divide both sides by 9.8 we get mass is equal to about 3.47kg.

is xenon a pure substance​

Answers

[tex]\large\huge\green{\sf{Yes}}[/tex]

a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. a.) What is the speed of the center of gravity when the rod reaches its lowest position? b.) What is the tangential speed of the lowest point of the rod when the rod reaches its lowest position?

Answers

Answer:

Explanation:

Potential energy gets converted to rotational kinetic energy

a)         ½Iω² = mgh

½(mL²/3)ω² = mgL/2

       (L/3)ω² = g

              ω = [tex]\sqrt{3g/L}[/tex]

       v(CG) = (L/2) [tex]\sqrt{3g/L}[/tex]

Not sure if you wanted angular speed or tangential speed of the CG so I gave both.

b) v = =  L [tex]\sqrt{3g/L}[/tex]

4. Protons and neutrons are held together to form this _________

Answers

Answer:

strong nuclear force.

Explanation:

hope this helps you!!

This is physics and it says collision and elastic/inelastic i need help

Answers

The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.

In this problem, we have to apply the law of conservation of linear momentum. Note that;

Momentum before collision = Momentum after collision

m1u1 + m2u2 = m1v1 + m2v2

(2150 × 10) + (3250u1) = (2150 + 3250)5.22

21500 + 3250u1 = 5400 × 5.22

3250u1  = 28188 - 21500

u1 = 28188 - 21500/3250

u1 = 2.1 m/s

2) Again from the principle of conservation of linear momentum;

(0.40 × 3.5) + (0.60 × 0) = (0.40  × 0.70) + (0.60 × v2)

1.4 = 0.28 + 0.60v2

1.4 - 0.28 =  0.60v2

v2 = 1.87 cm/s

Using;

s = 1/2 ( u + v)t

s = 1/2(0 + 1.87) × 5

s = 4.7 cm

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I need help with parts a and B of this question

Answers

Answer:

Explanation:

Let x be the leaf spring compression distance

F = kx

5.20 x 10⁵ = (5.45 x 10⁵)x + (3.80 x 10⁵)(x - 0.500)

5.20 x 10⁵ = (5.45 x 10⁵)x + (3.80 x 10⁵)(x) - 1.90 x 10⁵

7.10 x 10⁵ = (9.25 x 10⁵)x

x = 0.76756...

x = 0.768 m

W = ½(5.45 x 10⁵)0.768² +  ½(3.80 x 10⁵)(0.768 - 0.500)²

W = 174,148.648648...

W = 174 KJ

What happens to the gravitational force between two objects if the distance between them triples?

A. The force increases by a factor of 9

B. The force decreases by a factor of 9

C. The force decreases by a factor of 3

D. The force increases by a factor of 3

Answers

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. ... If the separation distance between any two objects is tripled (increased by a factor of 3), then the force of gravitational attraction is decreased by a factor of 9 (3 raised to the second power).

So the answer you are looking for is B.
The answer will be B.

Gravity obeys something called the inverse square law.

This means if distance increases by a factor of x, the force of gravity will decrease by a factor of x^2

For example, if distance increases by 2, the force would be 4 times weaker

If Distance increases by a factor of 3, the force will be 9 times weaker


Two Blocks are connected by a massless rope over a massless,
frictionless pulley, as shown in the figure. The mass of block 2
is m2 = 10.1 kg, and the coefficient of kinetic friction
between block 2 and the incline is Mk = 0.200. The angle 0 of
the incline is 27.5º. If block 2 is moving up the incline at
constant speed, what is the mass mi of block 1?

Answers

The mass of block 1 will be 1.99 kg.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.

What is tension force?

The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.

Given that,

Mass of block 1=? kg

The coefficient of the kinetic friction,μ=0.200

Now consider the weight component in the uphill direction.The weight is found as;

[tex]\rm W=m_1gsin \theta[/tex]

The force is balanced in the vertical direction as;

[tex]\rm T=F_f-W[/tex]

When the force of friction is;

[tex]\rm F_F=\mu_k N[/tex]

[tex]\rm F_f=(m_1 gcos \theta)[/tex]

Substitute the value in the vertical balanced equation;

[tex]\rm T=m_1gsin(27.5)^0-\mu_kmgcos27.5^0[/tex]

[tex]\rm T-m_2g=0\\\\T=m_2g[/tex]

[tex]\rm (10.1) g=m_1g(0.699-0.2 \times(-0.714) ) \\\\ (10.1) g=m_1g (0.699+0.1428) \\\\\ (10.1) g= m_1 \times 0.8418 \\\\ m_1 =11.99 \ kg[/tex]

Hence the mass of block 1 will be 1.99 kg.

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