Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 2.857 m/s/s. The acceleration lasts for 15.5 s. What is the final speed of the car and how much ground does it cover during this acceleration?

Answers

Answer 1

Answer:

Vf = 44.56 [m/s]

Explanation:

In order to solve this problem we must use the following expression of kinematics.

[tex]v_{f} = v_{i}+(a*t)[/tex]

where:

Vf = final velocity [m/s]

Vi = initial velocity = 0

a = acceleration = 2.857 [m/s^2]

t = time = 15.5 [s]

Note: the initial velocity is equal to zero as the car begins its movement from rest, or with an initial velocity equal to zero.

Vf = 0 + (2.875*15.5)

Vf = 44.56 [m/s]


Related Questions

Why is it important to warm up your muscles?
Warming up the muscles increases lactic acid build up
Warming up the muscles increases temperature and flexibility of your muscles Warming up the muscles increases your risk of injury
All of the above

Answers

Answer: Warming up the muscles increases temperature and flexibility of your muscles
Hey :)

The answer would be

B. Warming up the muscles increase temperature and flexibility of your muscles.

Because, lactic acid buildup is from overexerting yourself when working out, and warming up muscles does not increase your risk in injury, rather, it decreases your risk of injury

Hope this helps!

Which statement describes S waves? A. They move side to side.B.They arrive betore P waves. C. They travel through liquids D. They occur at Earth's surtace.​

Answers

Answer: they move side to side

Explanation:

i took the test

Select the correct answer.
Which scientist discovered the relationship between electricity and magnetism?
OA.
Nikola Tesla
OB
Alessandro Volta
OC.
Hans Christian Oersted
OD
Isaac Newton

Answers

Answer:

Hans Christian Oersted

Explanation:

A 10 [kg] object is dropped from rest. a. How far will it drop in 2 [s]?

Answers

Answer:

y=0.5 g t^2

=0.5*10*2^2

=20 m

Calculate the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I. (Note the loop and wire are not in electrical contact.)

Answers

Answer:

The magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I is [tex]\pi[/tex] or 3.14

Explanation:

The magnetic field at center of a circular loop is given by

[tex]B = \frac{\mu _{o}I }{2R}[/tex]

Where B is the magnetic field

[tex]\mu _{o}[/tex] is the free space permeability constant ( [tex]\mu _{o}[/tex] = 4π × 10⁻⁷ N/A²)

[tex]I[/tex] is the current

and [tex]R[/tex] is the radius

For the magnetic field of a long straight wire, it is given by

[tex]B = \frac{\mu _{o}I }{2\pi R}[/tex]

Where B is the magnetic field

[tex]\mu _{o}[/tex] is the free space permeability constant ( [tex]\mu _{o}[/tex] = 4π × 10⁻⁷ N/A²)

[tex]I[/tex] is the current

and [tex]R[/tex] is the distance from the wire

Then, to calculate the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I, that will be

[tex]\frac{\mu _{o}I }{2R} \div \frac{\mu _{o}I }{2\pi R}[/tex]

= [tex]\frac{\mu _{o}I }{2R} \times \frac{2\pi R }{\mu _{o}I}[/tex]

= [tex]\pi[/tex]

(NOTE: [tex]\pi[/tex] = 3.14)

Hence, the magnetic field at the center of a circular current loop of radius R divided by the magnetic field at a distance R away from a very long straight wire carrying the same current value I is [tex]\pi[/tex] or 3.14.

PLEASE HLEP ME ASAP

Tom, a 13 year old boy, wants to train for a 5k Fun Run that his school is organizing. What area of fitness should Tom focus on? What activities should he do to improve this area?

Answers

Answer:

cardiovascular endurance

Explanation:

In order for Tom to successfully complete the marathon, he should focus on being able to maintain a constant pace without tiring easily.

a hole is punched at a height h in the side of a container of height h0. The container is full of water. If the water is to shoot as far as possible horizontally, (a) how far from the bottom of the container should the hole be punched? (b) Neglecting frictional losses, how far (initially) from the side of the container will the water land?​

Answers

a) the hole needs to be as close to the bottom as possible for maximum horizontal projection.

b) not sure there is enough information to answer that. It depends where the hole is.

Add the following vectors:

(a) (12, 5) + (6, 3)
(b) (−3, 8) + (6, −2)
(c) (3, 8, −7) + (7, 2, 17)
(d) (a, b, c) + (d, e, f)

Answers

Answer:

a. (18,8)

b. (3, 6)

c. (10, 10, 10)

d. (a+d, b+e, c+f)

Give me brainliest for this... Thanks

Can someone please help me

Answers

Answer:

grams= 4.8e-5

kilograms= 4.8e-8

centigrams= 0.0048

dekagrams= 4.8e-6

Explanation:

Social science is a branch of science that deals with the nature of
A. human society
B. the physical world
C. animal societies
D. the medical field

Answers

The answer is A. Human Society.

Answer:

A. human society

Explanation:

Atoms tend to be stable and nonreactive if they have six valence electrons.
Select one:
O True
O False

Answers

Answer:

True

Explanation:

What is the takeoff distance for an airplane on a runway at
300km/hr with acceleration of 1 m/s 2

Answers

Answer:

x = 3472 [m]

Explanation:

To solve this problem we must use the following expression of kinematics:

[tex]v_{f}^{2}= v_{i}^{2}+(2*a*x)\\[/tex]

where:

Vi = initial velocity = 0

Vf = final velocity = 300 [km/h]

a = acceleration = 1 [m/s^2]

x = takeoff distance [m]

Note: The initial velocity is equal to zero as avion starts its movement from rest.

Now we have to convert units of kilometers per hour to meters per second, the final velocity.

[tex]300 [\frac{km}{h}]*[\frac{1h}{3600s} ]*\frac{1000m}{1km} \\= 83.33 [\frac{m}{s} ][/tex]

(83.33)^2 = 0 + (2*1*x)

2*x = 6943.88

x = 3472 [m]

2
A crane has an arm to which is attached a frictionless pulley. A cable passes over the pulley and
supports a load of 10 kN as shown.
frictionless pulley
104N
30
30°
crane arm
load
10 KN
The crane arm exerts a force F on the pulley.
What is the value of F?
C
8.7 KN
10 KN
D
A
17 kN
B
5.0 KN

Answers

Answer:

17kN

Explanation:

The force of 10 kN in the cable and the load of 10 kN both have a component along the crane arm.

The force F exerted by the crane arm on the pulley is the sum of these 2 components.

Component of force in cable along the crane arm = 10 cos 30°

Component of load along the crane arm = 10 cos 30°

Considering the forces on the pulley,

F = 10 cos 30° + 10 cos 30°

Force F = 17.32 kN = 17 kN

A shot-putter exerts an unbalanced force of 128 N on a shot giving it an acceleration of 19
m/s2. What is the mass of the shot?

Answers

Answer:

128 is the ans cuz N is also lnown as mass

Explanation:

128

A bus rolls to a stop along a horizontal road without the driver applying the brakes,
What answer choice BEST explains why this occurred?
A. The natural state of the bus is not to be in motion.
B. The bus ran out of momentum.
C. The force of gravity slowed the bus until it stopped.
D. The opposing force of friction stopped the train.

Answers

Answer:

should be d because friction allows things to go faster or slower

It’s opinion a I’m a 100%

a leaky faucet is dripping at a rate of 1 pint per hour how many hours to fill 10 gallons

Answers

Answer:

80 hours :)))))))))))))

Erica is working in the lab. She wants to remove the fine dust particles suspended in a sample of oil. Which method is she most likely to u

Answers

Answer:  Option A is your answer

Explanation:

1.What is the mass of an object that weighs 98 N?​

Answers

Answer:

10 kg

Explanation:

Answer: 10kg




10 kg newton

A projectile is fired into the air at an angle of 50° above ground level and hits a target downrange. Neglecting air drag, it will also hit the target if fired at the same speed at an angle of

Answers

Answer:

18 m/s

Explanation:

Range of a projectile on level ground is:

R = v₀² sin(2θ) / g

14.3 m = v₀² sin(2×13°) / 9.8 m/s²

v₀ = 17.9 m/s

Rounded to two significant figures, the launch speed was 18 m/s.

If the bullet is launched at an angle of 50 degrees above ground level, the target will be struck. The angle remains the same. The launch angle obtained is 50 degrees.

Given:

The initial shot was fired at an angle of 50 degrees above ground.

The projectile's starting velocity (v) and magnitude of velocity will remain constant if it is shot at the same pace.

Let the angle of the projectile is x,

The horizontal component of velocity can be calculated as follows:

[tex]v(x) = v * cos(x)[/tex]

We can write:

since the horizontal part of velocity remains constant:

[tex]v(x1) = v(x2)[/tex]

[tex]cos(50) = cos(x)[/tex]

[tex]50 = x[/tex]

Therefore, if the projectile is launched at the same speed at a 50° angle above ground level, it will strike the target.

To know more about the projectile:

https://brainly.com/question/13388411

#SPJ4

The correct response is given when the angle is asked a question.

As part of a carnival game, a 5.00 kg target is freely hanging from a very long and very light wire. Contestants can use one of two 1.5 kg balls to try to hit the target and deflect it high enough to win a prize. Ball A will have an elastic collision and bounce back toward you while ball B will have a nearly perfectly inelastic collision, but rather than sticking to the target, the ball will just drop straight downward to the ground after the collision. You can throw each ball with a velocity of 12 m/s. You are the first to try the game and which ball should you throw? Calculate the expected height the target will reach after each is thrown.

Answers

Answer:

Height ball A will deflect the target is 2.65 m

Height ball A will deflectthe target is 2.2 m

Ball A, will deflect the target to greater height, thus i will throw ball A

Explanation:

Given;

mass of the target, m₁ = 5.00 kg

mass of the ball, m₂ = 1.5 kg

initial velocity of each throw, u = 12 m/s

Throwing ball A; apply the principle of conservation linear momentum for elastic collision;

m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂

The initial velocity of the target, u₁ = 0

The ball bounced back at the same speed, v₂ = -12 m/s

the velocity of the target after collision, = v₁

0 + 1.5 x 12 = v₁(5) + (-12 x 1.5)

18 = 5v₁ - 18

18 + 18 = 5v₁

36 = 5v₁

v₁ = 36/5

v₁ = 7.2 m/s

The vertical height reached by the target is given by;

v₁² = u₁² + 2gh

v₁² = 0 + 2gh

v₁² = 2gh

h = v₁² / 2g

h = (7.2)² / (2 x 9.8)

h = 2.65 m

Throwing ball B; apply the principle of conservation energy for the inelastic collision;

the kinetic energy of the ball will be converted to the potential energy of the target.

¹/₂m₁u₂² = mgh

¹/₂(1.5)(12)² = (5 x 9.8)h

108 = 49h

h = 108 / 49

h = 2.2 m

Ball A will deflect  the target to greater height, thus i will throw ball A.

Can you help me please! Love you​

Answers

Answer:

I think its B. wait...U love ME?????

Which of the following choices is not part of the scientific method

Answers

Answer:

What are the choices?

Explanation:

What is called mass per unit

volume of a substances? | ककसी पदाथष के

द्रव्यमान प्रतत इकाई आयतन को क्या कहते हैं?

A : Mass | द्रव्यमान

B : Weight | भार

C : Density | घनत्व

D : Volume | आयतन

Ans:
C : Density | घनत्व​

Answers

Answer:

Density

Explanation:

The definition of density of a substance is the quotient between its mass and its volume. Such results in the mass per unit of volume.

I need helpppppppppppppppppppppppppppppppppppppppppppppp

Answers

Answer:u

Laa             c

Explanation:


A ball is thrown upward in the air with an initial velocity of 40 m/s. How long does it take to
reach back to the point it was thrown from?

Answers

Answer:

You need the definition of acceleration (a=Vf-Vi/t) and 1 equation of linear motion (deltaX = Vi×t + 1/2×a×t^2). Since you know a is constant (gravity) and you know your initial Vi to be 40 m/s and your final velocity Vf to be zero (maximum height), then you can use thhe definition of acceleration to find time.

-9.81m/s^2 = (0-40m/s)/t

t = (-40)/(-9.81) s

t = 4.077s

Now that you have time, you should know all but deltaX in the equation of linear motion.

dX = (40m/s)(4.077s) + (1/2)(-9.81m/s^2)(4.077s)^2

dX = (163.099m) — (81.549m)

dX = 81.55m

the graph below represents the relationship between velocity and time for an object moving along a straight line.

What is the magnitude of the object's acceleration?

1. 5.0 m/s^2

2. 8.0 m/s^2

3. 10. m/s^2

4. 20. m/s^2​

Answers

Answer:

10. m/s^2

Explanation:

40-(-10)= 50

50/5 =10

This question can simply be answered by using the slope of the velocity-time graph.

The correct answer for the magnitude of the acceleration is "3. 10 m/s²".

Since the graph between the velocity and the time is in the form of a straight line. Hence, the acceleration can be calculated as the slope of this line. We will use the first and last point of the line to calculate the slope (acceleration):

[tex]Acceleration = Slope = \frac{v_2 - v_1}{t_2-t_1}[/tex]

where,

v₁ = initial velocity from the graph = - 10 m/s

v₂ = final velocity from the graph = 40 m/s

t₁ = initial time from the graph = 0 s

t₂ = final time from the graph = 5 s

Therefore,

[tex]Acceleration = \frac{40\ m/s\ -\ (-\ 10\ m/s)}{5\ s\ -\ 0\ s}[/tex]

Acceleration = 10 m/s²

The attached picture shows the types of a velocity-time graph.

Learn more about the velocity-time graph here:

https://brainly.com/question/11290682?referrer=searchResults

What is the formula for speed?

गतत का सूत्र क्या है?

A : Distance covered/Time | तय की गई

दरूी/समय

B : Change in velocity/Time | वेग में

पररवतषन/समय

C : Distance in definite direction /Time |

तनजचचत टदशा में दरूी/समय

D : Change in momentum/Time | संवेग में

पररवतषन/समय
Ans:
A : Distance covered/Time | तय की गई

दरूी/समय

Answers

Explanation:

Formula for speed = Distance covered/Time

Therefore,

Option A is correct.

Answer:

a.distance÷timetaken

Explanation:

The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).

Which of the following would be useful to show the growth rate of plants at varying temperatures?


bar graph


circle graph


line graph


frequency table

Answers

Answer:

line graph

Explanation:

Ik im late but i hope this helps lol

A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed

Answers

Complete Question

A sled of mass m is coasting at a constant velocity on the ice covered surface of a lake. Three birds, with a combined mass 0.5m, gently land at the same time on the sled. The sled and birds continue sliding along the original direction of motion. How does the kinetic energy of the sled and birds compare with the initial kinetic energy of the sled before the birds landed?

a) The final kinetic energy is one half of the initial kinetic energy.

b) The final kinetic energy is one third of the initial kinetic energy.

c) The final kinetic energy is one quarter of the initial kinetic energy.

d) The final kinetic energy is one ninth of the initial kinetic energy.

e) The final kinetic energy is equal to the initial kinetic energy

Answer:

The correct option is b

Explanation:

From the question we are told that

   The mass of the sled is  m  

   The  mass of the three birds is  [tex]m_b = 0.5 m[/tex]

Generally from the law of momentum conservation,

  The initial momentum of the boat  =  final  momentum of the boat

Hence

     [tex]m * v = (m_b+ m) * v_2[/tex]

=>  [tex]m v = (0.5m+ m) * v_2[/tex]

=>  [tex]m v = 1.5m* v_2[/tex]

Here v is the velocity of the boat before the birds landed and [tex]v_2[/tex] is the velocity after the birds landed

So

       [tex]v_2 = \frac{2}{3} v[/tex]

Generally the initial  velocity of the boat before the birds landed is mathematically represented as

      [tex]K_i = \frac{1}{2} * m * v ^2[/tex]

Generally the final  velocity of the boat when the birds have landed is mathematically represented as

      [tex]K_f = \frac{1}{2} * (m_b + m ) * v_2^2[/tex]

=>   [tex]K_f = \frac{1}{2} * (0.5m + m ) * (\frac{2}{3} v)^2[/tex]

=>  [tex]K_f = \frac{1}{2} * 1.5m * (\frac{2}{3} v)^2[/tex]

=>  [tex]K_f = \frac{1}{2} * 1.5m * \frac{4}{9} v^2[/tex]

=>  [tex]K_f = \frac{1}{2}m v^2 * \frac{1}{3}[/tex]

Comparing this equation with that of the initial  kinetic energy

=>  [tex]K_f = \frac{1}{3} K_i[/tex]      

The kinetic energy of the sled and birds is one-third the initial kinetic energy of the sled before the birds landed.

What is kinetic energy?

When an object is in motion, then the energy possessed by the object at the state of motion is known as its kinetic energy.

Given data -

The combined mass of three birds is, M = 0.5m.

Let m be the mass of the sled. Then applying the conservation of linear momentum as,

The initial momentum of the boat  =  final  momentum of the boat

mv = (M + m)v'

v is the velocity of the boat before the birds landed and v' is the velocity after the birds landed.

mv = (0.5m + m)v'

mv = 1.5mv'

v = 3/2 v'

or

v' =2/3v

The initial kinetic energy of the boat before the birds landed is mathematically represented as

[tex]K_{1}= \dfrac{1}{2}mv^{2}[/tex]

And the final kinetic energy of the boat when the birds have landed is mathematically represented as,

[tex]K_{2}=\dfrac{1}{2}(M+m)v'^{2}\\\\K_{2}=\dfrac{1}{2}(1.5m) \times 4/9v^{2}\\\\K_{2}=\dfrac{1}{3} \times \dfrac{1}{2}mv^{2}\\\\K_{2}=\dfrac{1}{3} \times K_{1}[/tex]

Thus, we can conclude that the kinetic energy of the sled and birds is one-third the initial kinetic energy of the sled before the birds landed.

Learn more about kinetic energy here:

https://brainly.com/question/999862

what are Newton's 3 laws

Answers

Answer:

Newton 3 laws

Explanation:

1.object will not change its motion unless a force acts on it . 2.the force on an object is equal to it's time mass on acceleration. 3. When 2 o jecys interact they apply forces to each other of equal.

Answer- In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Hope this helps:)
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