A race car makes one lap around a track with a radius of 41 meters in 8 seconds. What is the velocity?

Answers

Answer 1

Answer:

31.21 m/s

Explanation:

radius=41m

time = 8 sec

circumference=2πr

from formula

circumference=2×22/7×41

=257.71m

now

velocity=total distance( circumference)/total time

=257.71/8

= 32.21m/s

Answer 2

The velocity of race car makes one lap around a track with a radius of 41 meters in 8 seconds is 32.19 m/s

What is velocity ?

Velocity is "rate of change of displacement with respect to time".

i.e. v= dx/dt

it is also defined as displacement over time. i.e. v=Displacement/Time.

Velocity shows how much distance can be covered in unit time. It's SI unit is m/s. It is vector quantity ( having both direction as well as momentum). where displacement is distance from mean position.

Given,

Radius of the track r= 41m

time taken by car to complete one lap T = 8s

velocity v = ?

we know that ω = 2π÷T

ω = 2×3.14 ÷ 8

ω = 0.785 rad/s

v=rω

v= 41×0.785

v= 32.185 ≅32.19 m/s

Hence velocity of the car is 32.19 m/s

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Related Questions

an object has an mass of 15 kg and is falling at a rate of 2.0 m/s what is the momentum?

Answers

Answer:

30 kg.m/s

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 15 × 2

We have the final answer as

30 kg.m/s

Hope this helps you

A man speeding at 40m/s decides to outrun the cops and starts to
accelerate at a rate of 2.5m/s2 for 12 seconds. What is the criminal's new
speed?

Answers

Answer:

70 m/s.

Explanation:

Given that,

Initial speed, u = 40 m/s

Acceleration = 2.5 m/s²

Time, t = 12 s

We need to find criminal's new  speed. Let it is v. Using equation of motion to find it as follows :

v = u +at

Substitute all the values

v = 40 + 2.5(12)

v = 70 m/s

So, the new speed is 70 m/s.

a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy

What is the squirrels mass

Answers

Answer:

yeet yeet yeet yeet

Explanation:

Kinetic energy (K.E):-

So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.

A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.

What is the squirrel’s mass?

Answer: 0.51 kg

5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley to a 0.0150 kg mass.

Draw a free body diagram showing all the forces acting on each object
Calculate the acceleration of the masses
Calculate the tension force in the cord
How long will it take the block to get to go 1.2 m to the edge of the table.
What will the velocity be as soon as it gets to the edge?

Answers

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

Answer:

There's no answer I'm just taking points like you did me,  so thank you for your points I'll put them to good use ;)

Give one example where friction is useful. Can someone feed me with correct answer pls c:

Answers

Answer:

Friction is what holds your shoe to the ground. The friction present on the ice is very little, this is the reason why it is hard to walk on the slippery surface of the ice.

Explanation:

Keeping traction on a race track

A 250-kg moose stands in the middle of the railroad tracks in Sweden, frozen by the lights of an oncoming 10,000kg train traveling at 20m/s. Even though the engineer attempted in vain to slow the train down in time to avoid hitting the moose, the moose rides down the remaining track sitting on the train’s cowcatcher. What is the final velocity of the train and moose after the collision?
(Momentum & Impulse)

Answers

Answer:

The final velocity of the train and the moose after collision is approximately 19.51 m/s

Explanation:

The given mass of the moose, m₁ = 250 kg

The velocity of the moose, v₁ = 0

The mass of the oncoming train, m₂ = 10,000 kg

The velocity of the train, v₂ = 20 m/s

The velocity of the moose and the train after collision = v₃

By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Therefore, by substitution, we have;

250×0 + 10,000× 20 = (10,000 + 250) × v₃

200,000 = 10,250 × v₃

v₃ = 200,000/10,250 ≈ 19.51 m/s

The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s

A friend is coming to Tim’s house to study after school. What directions would Tim give for reaching his house from the entrance of the school?
(I WILL GIVE BRAINLIEST)

Answers

Answer:

go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school

An object that falls and accelerates solely as a result of gravity is said to be in
(2 points)
A. terminal velocity
B. free fall
C. air resistance
D. terminal acceleration

Answers

I think the answer is c

9. A student notices that wearing darker colors in sunlight makes him feel warmer, so he decides to conduct an experiment. He takes five pieces of different
colored cloth and wraps
each one around a water bottle. He then places all five bottles in direct sunlight and measures the temperature of the water in each bottle an hour later
What is the dependent variable in this experiment?
O the time he leaves it in the sunlight
O the amount of water in each bottle
O the color of the cloth
O the temperature of the water

Answers

Answer: 4

Explanation:

The dependent variable is the temperature of the water.

A golf ball is sitting on a tee. The ball is struck with a golf club and flies
through the air. How does the force on the club compare with the force on the
ball when momentum is transferred between the club and ball?

Answers

Answer:

c i kn now it is

Explanation:

1. While riding a chairlift, a 55-kg skier is raised a vertical distance of 370 m. What is the total change in the skier's gravitational potential energy

Answers

Answer:

199,430Joules

Explanation:

Gravitational potential energy is the energy possessed by a falling object due to virtue of its position.

gravitational potential energy = mass * acceleration due to gravity * height

gravitational potential energy = 55 * 9.8 * 370

gravitational potential energy = 539 * 370

gravitational potential energy = 199,430

Hence the total change in the skier's gravitational potential energy is 199,430Joules

ANSWER:

99,430 Joules (in multiple choice question sometimes appear as 200,000J)

Explanation:

Gravitational potential energy is the energy possessed by a falling object due to virtue of its position.

gravitational potential energy = mass * acceleration due to gravity * height

gravitational potential energy = 55 * 9.8 * 370

gravitational potential energy = 539 * 370

gravitational potential energy = 199,430

Hence the total change in the skier's gravitational potential energy is 199,430 Joules (or 200,00J)

please heart!VVV          or stars :D vvv

Jerry is pushing a 50-kg box across a moth floor with an acceleration of 0.6 m/s2. What force is he applying to the box? *

83.3 N
0.012 N
0
30 N

Answers

Answer:

30 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 50 × 0.6

We have the final answer as

30 N

Hope this helps you

How much would a pair of 0.5 kg shoes weigh on Earth? (Include units in
your answer) *

Answers

Answer:

1.1 lbs

Explanation:

To convert kg to lbs you multiply kilograms by 2.2. So 0.5kg × 2.2 equals to 1.1 lbs

help please asap due 20 minutes please help me ​

Answers

did you turn it in yet?

Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente hacia adelante para que cuando ésta toque la superficie del agua haga varios "saltos" sobre el agua. Durante un juego de estos, un desocupado nota que una de las piedras que arroja se demora 0,4 s en tocar la superficie del agua y la toca a 2,5 m de la orilla del lago, desde donde fue lanzada. Encuentre: a) La altura de la que fue lanzada la piedra. b) La velocidad con la que fue lanzada.

Answers

Answer:

a) La piedra es lanzada desde una altura de 0,785 metros.

b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Explanation:

a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:

[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)

Donde:

[tex]y[/tex] - Altura final, medida en metros.

[tex]y_{o}[/tex] - Altura inicial, medida en metros.

[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.

[tex]t[/tex] - Tiempo, medido en segundos.

[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:

[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]

[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]

[tex]y_{o} = 0,785\,m[/tex]

La piedra es lanzada desde una altura de 0,785 metros.

b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:

[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)

Donde:

[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.

[tex]t[/tex] - Tiempo, medido en segundos.

Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:

[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]

[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]

La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which water hits a wall at 8 m away from the hose (consider that acceleration due to gravity =9.8 m/s2)

Answers

Answer:

The water hits the wall at a height of 5.38 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

The object describes a parabolic path given by the equation:

[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]

Where:

y   = vertical displacement

x   = horizontal displacement

θ   = Elevation angle

vo = Initial speed

The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.

The height at which the water hits a wall located at x=8 m from the hose is:

[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]

Calculating:

y = 5.38 m

The water hits the wall at a height of 5.38 m

Can u anser 5,6 on the picture

Answers

Answer: Number 6 is Periods

Explanation:

NEED AWNSER NOW! WILL MARK BRAINLY! Which term is defined as the ratio of the speed of light in a vacuum to the speed of light in the material it is passing through?

index of reflection

index of refraction

angle of reflection

angle of incidence

Answers

Answer:

Index of refraction.

Answer:

index of refraction

Explanation:

I just took the k12 quiz.

A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring

Answers

The forces constant (spring constant) of the spring will be 41.09 N/m.

What is spring force?

The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;

F = kx

The given data in the problem is;

F is the spring force = 30 N

K is the spring constant= ?

x is the displacement of spring = 0.73 m

The spring constant is;

K =F/x

K=30/0.73

K=41.09 N/m

Hence the force constant (spring constant) of the spring will be 41.09 N/m.

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A fish swimming at a rate of .6 m/s notices a huge shark. Three seconds later, the fish is swimming at a speed of 3 m/s. What is the fish's acceleration?

0.8 m/s/s
-0.8 m/s/s
12.5 m/s/s
-12.5 m/s/s

Answers

Answer:

C

Explanation:

???

i think

In a place covered by shadow of cloud sun cannot be seen . Explain with reasons .

Answers

Answer:

Because even though our eyes have a huge dynamic range (ability to pick out details in sharply lit and lesser lit areas simultaneously) than any camera, there's a limit.

When there's strong sunlight, your pupils contract and let less light in, which makes the shadows look darker.

When it's cloudy, your pupils widen and let more light in, which makes the shadows look less dark.

Do some experiments with a camera and you'll soon get the hang of it.

NOTE: Also test HDR (high dynamic range) photography, where the camera takes three or more pictures in quick succession, with different exposure settings, and combines them to get the most detail of both bright and dark areas. The result is more or less what we percieve.

Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero

Answers

Answer:

The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.

draw position time graph when speed is increasing​

Answers

Explanation:

We need to draw position-time graph when the speed is increasing.

The slope of position-time graph gives the speed of an object.

Position means distance covered.

When the speed of an object is increasing with time. It means it is moving with increasing speed.

The attached figure shows the position -time graph when speed is increasing​.

Which electron dot diagram shows the bonding between 2 chlorine atoms?2 dots then C l with 2 dots above and 1 dot below then 2 dots then 2 dots then C l with 2 dots above and 1 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 2 dots then C l with 2 dots above and 2 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 1 dot then C l with 2 dots above and 2 dots below then 2 dots.2 dots then C l with 2 dots above and 1 dot below then 3 dots then 3 dots then C l with 2 dots above and 1 dot below then 2 dots.

Answers

Answer:

It is B

Explanation:

Answer: 2nd answer

Explanation: took exam

A 0.15 kg ball is moving with a velocity of
35 m/s. Find the momentum of the ball.

Answers

Answer:

5.25 kg.m/s

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 0.15 × 35

We have the final answer as

5.25 kg.m/s

Hope this helps you

13. Austin rode his bike 10 m/s for two minutes. How far did he travel? A. 200 meters B. 1200 meters C. 1000 meters D. 20 meters​

Answers

Answer:

B. 1200

Explanation:

60 sec in one min in 2 min there will be 120 sec. 10x120=1200

The diagram shows a model of an atom. Who first proposed this model?
A. Bohr
B. Thomson
C. Rutherford
ОО
D. Dalton

Answers

A.Bohr

His model postulated the existence of energy levels or shells of electrons. Electrons could only be found in these specific energy levels; in other words, their energy was quantised, and couldn’t take just any value. Electrons could move between these energy levels but had to do so by either absorbing or emitting energy.

A. Bohr!

This answer is correct because I read the information.

show your work. john uses a 25N force to push a boulder off a cliff that is 312m tall. What is the work done on the boulder?​

Answers

Answer:7800

work=force x distance

Force in Newtons

Distance in Meters

Work in Joules

The magnitude of vector vector A is 84.9 m and it points in the +y axis direction. The magnitude of vector vector B is 195.0 m and it points at an angle of 41.0° counterclockwise from +x axis. The magnitude of vector vector C is 126.2 m and it points in the +x axis direction.

Answers

Solution:

The magnitude of A vector is 84.9 m in the positive y-axis direction.

So the X component of A =0

     the Y component of A = 84.9 m

Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.

So  the X component of B = B cos 41°

                                           = 195 x cos 41°

                                           = 195 x 0.75   = 146.25 m

   the Y component of B = B sin 41°

                                           = 195 x sin 41°

                                           = 195 x 0.65   = 126.75 m

Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.

So the X component of C =126.2 m

     the Y component of C = 0

Comparing all these, we get

1. B vector has the largest X component

2. B vector has the largest Y component

When two ocean plates come together, one ocean plate __________________
under the other, causing a chain of ________________ __________________
to form.

Answers

Answer:

A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.

Explanation:

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