If the velocity of a car changes from 0 meters per second (m/s) to 100 m/s in 10 seconds, what is the acceleration over that 10 second period?

Answers

Answer 1

Answer:

10m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 100m/s

Time taken  = 10s

Unknown:

Acceleration  = ?

Solution:

Acceleration is the rate of change of velocity with time.

  A = [tex]\frac{v - u}{t}[/tex]

v = final velocity

u = initial velocity

t = time taken

 So, insert the parameters and solve;

  A = [tex]\frac{100 - 0}{10}[/tex]   = 10m/s²


Related Questions

A 0.20 kg mass (m1) hangs vertically from a spring and an elongation of the spring of 9.50 cm (r1) is recorded. With a mass (m2) of 1.00 kg hanging on the spring, a second elongation (r2) of 12.00 cm is recorded. Calculate the spring constant k in Newtons per meter (N/m). (Note: The equilibrium position is not zero.)

Answers

Answer:

k=320N/m

Explanation:

Step one:

given data

Let the initial/equilibrum position be x

mass m1= 0.2kg

F1= 0.2*10= 2N

elongation e= 9.5cm= 0.095m

mass m2=1kg

F2=1*10= 10N

elongation e= 12cm= 0.12m

Step two:

From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F

F=ke

2=k(0.095-a)

2=0.095k-ka----------1

10=k(0.12-a)

10=0.12k-ka----------2

solving equation 1 and 2 simultaneously

 

   10=0.12k-ka----------2

-   2=0.095k-ka----------1

   8=0.025k-0

divide both side by 0.025

k=8/0.025

k=320N/m

Consider a compact car that is being driven
at 113 km/h. The acceleration of gravity is 9.8 m/s^2.
From what height would the car have to be
dropped to have the same kinetic energy?
Answer in units of m.

Answers

The answer is 2847.830m

According to the Law of Reflection, the angle of incidence the angle of reflection. O A. is greater than B. is less than C. equals D. is opposite from​

Answers

Answer:

C. Equals

Explanation:

Law of reflection Equals the angle of incidence

A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant of the spring?

Answers

Answer:

Force constant of the spring (k) = 24.07 N/m

Concept/Theory:

The period [tex] \sf (T_s) [/tex] of a spring-mass system is proportional to the square root of the mass (m) and inversely proportional to the square root of the force constant of the spring (k).

Equation of period:

[tex] \boxed{ \bf{T_s = 2 \pi \sqrt{\dfrac{m}{k}}}}[/tex]

Explanation:

Mass = 0.250 kg

Period = 0.640 s

By substituting values in the equation, we get:

[tex] \rm \longrightarrow 0.640 = 2 \pi \sqrt{\dfrac{0.250}{k}} \\ \\ \rm \longrightarrow 2 \times 3.14 \sqrt{ \dfrac{0.25 0}{k} } = 0.640 \\ \\ \rm \longrightarrow 6.28 \sqrt{ \dfrac{0.250}{k} } = 0.640 \\ \\ \rm \longrightarrow \sqrt{ \dfrac{0.250}{k} } = \frac{0.640}{6.28} \\ \\ \rm \longrightarrow \frac{0.250}{k} = { \bigg(\frac{0.640}{6.28} \bigg) }^{2} \\ \\ \rm \longrightarrow \frac{k}{0.250} = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \\ \\ \rm \longrightarrow k = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \times 0.250 \\ \\ \rm \longrightarrow k = 24.07 \: N/m[/tex]

The force constant of the spring is approximately 24.038 newtons per meter.

As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant ([tex]k[/tex]), in newtons per meter, from the equation of the Period ([tex]T[/tex]), in seconds, which is described below:

[tex]T = 2\pi\cdot \sqrt{\frac{m}{k} }[/tex] (1)

Where [tex]m[/tex] is the mass of the moving element, in kilograms.

If we know that [tex]T = 0.640\,s[/tex] and [tex]m = 0.250\,kg[/tex], then the spring constant of the spring is:

[tex]0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }[/tex]

[tex]\sqrt{\frac{0.250}{k} } \approx 0.102[/tex]

[tex]\frac{0.250}{k} \approx 0.0104[/tex]

[tex]k \approx 24.038\,\frac{N}{m}[/tex]

The force constant of the spring is approximately 24.038 newtons per meter.

Please see this question related to Simple Harmonic Motion for further details: https://brainly.com/question/17315536

A cathode ray tube is made of glass with a small amount of some kind of gas in it. It has metal electrodes at each end to pick up an electric current. The electrodes are named "positive” and "negative." What are one of the main uses of this device ? to speed up the flow of current flowing through a wire to examine a beam of charged particles to magnetize different atoms within a space to change the strength of a proton within an atom

Answers

One of the main uses of this device is to examine a beam of charged particles

The cathode ray tube is a device used to determine the charge flowing in a gas. When an electric field is set up with the help of metal electrodes, the cathode ray tends to bend towards the positive electrode.

Since the cathode ray bends towards the electrodes, it implies that it has a charge and the electrodes present help us determine the charge of the beam of charged particles and thus examine the beam of charged particles.

So, one of the main uses of this device is to examine a beam of charged particles.

Learn more about cathode ray tube here:

https://brainly.com/question/17485156

Answer:

B

Explanation:

What is the relationship between resistance and current in a circuit with no change in voltage?
A. Current and resistance must be equal in a circuit.
B. A circuit that has more resistance will have smaller current.
C. Current does not depend on resistance in a circuit.
D. A circuit that has more resistance will have a greater current.

Answers

Answer:

A

Explanation:

the distance from Alex home to his school is 1km and 560 cm. what is this distance?​

Answers

Answer:

Required Answer:-1meter=100cm1km=1000m

[tex]{:}\longrightarrow[/tex][tex]\sf 1km=1000×100cm [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf 1km=100000cm [/tex]

Total distance

[tex]{:}\longrightarrow[/tex][tex]\sf 100000+560=1000560cm [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf 1000.56m[/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf 1.560km [/tex]

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A student releases a marble from the top of a ramp. The marble increases speed
while on the ramp then continues across the floor. The marble travels a total of
150cm in 4.80s.
What was the marble's final speed?

Answers

Explanation:

the formula of speed is distance traveled by time it work

Answer:

31.25 cm/sec

__________________________________________________________

Explanation:

We are given:

Distance travelled = 150 cm

Time taken = 4.8 seconds

Final Speed of the Marble:

Speed of the marble = Distance travelled / Time taken

Speed of the marble = 150 / 4.8

Speed of the marble = 31.25 cm/sec

A kid runs and slides down a slip-n-slide. Once the kid hits the slide they have 200N of friction force acting on them, and they decelerate at 2.5m/s. What is the mass of the kid?

Answers

Mass (kg) = force (N) / acceleration (m/s).

200/2.5 = 80



What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?

PLEASE SHOW WORK.

Answers

Answer:

The answer is 5.4 kg

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{135}{25} = \frac{27}{5} \\ [/tex]

We have the final answer as

5.4 kg

Hope this helps you

A child and sled with a combined mass of 54.8

kg slide down a frictionless hill that is 11.5 m

high at an angle of 36 degrees

from horizontal.

The acceleration of gravity is 9.81 m/s^2.

If the sled starts from rest, what is its speed

at the bottom of the hill?

Answer in units of m/s.

Answers

Answer:

15.02 m/s.

Explanation:

Given that the height of the hill, h= 11.5 m.

Combined mass, m= 54.8 kg

The initial velocity of the combined mass, u=0

Acceleration due to gravity, [tex]g = 9.81 m/s^2[/tex].

Angle of the path the horizontal, [tex]\theta = 36[/tex] degree.

Let A be the initial position and B be the final position of the sled as shown in the figure.

The path is frictionless so the drag force =0

The gravitational force acting on the combined mass in the downward direction, [tex]F= mg\cdots(i)[/tex]

The component of force acting in the direction of motion = [tex]F\sin \theta.[/tex]

Let [tex]a[/tex] be the acceleration of the combined mass, m, So,

[tex]F\sin \theta= ma[/tex]

[tex]\Rightarrow mg \sin \theta= ma[/tex] [ from equation (i)]

[tex]\Rightarrow a = g \sin \theta \cdots(ii).[/tex]

Let v be the final velocity of the combined mass.

Now, by using the equation of motion,

[tex]v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)[/tex]

Here, s is the displacement in the direction of motion,

So, s= AB

Now, in the right-angled triangle ABO,

[tex]\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)[/tex]

Now,  from equations (ii), (iii) and (iv), we have

[tex]v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}[/tex]

By using the given values, we have

[tex]v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s[/tex]

Hence, the speed of the combined mass at the bottom = 15.02 m/s.

The speed  of the sled at the bottom of the hill is [tex]15.02m/s[/tex]

The speed of sled is calculated by using Newton's law of motion,

                   [tex]v^{2} =u^{2} +2gh[/tex]

where u is initial velocity, v is final velocity , g is acceleration due to gravity and h is height.

Given that, [tex]u = 0, g = 9.81m/s^{2}[/tex] and [tex]h = 11.5 m[/tex]

Substitute values in above equation.

     [tex]v^{2}=0^{2}+2*9.81*11.5\\\\v^{2}=225.63\\\\v=\sqrt{225.63}=15.02m/s[/tex]

Learn more:

https://brainly.com/question/19030143

a car with a mass of 1200 kg travels a distance of 150 M as it moves from one stoplight to the next at its fastest the car travels at 22 m per second what is its kinetic energy at this point ​

Answers

Answer:

ke = 1/2mv^2

    = 1/2 * 1200 * 22^2

    = 600 * 484

    = 290400 J

Hope it helped u,

pls mark this as brainliest

and put thanks

^_^

Answer:

The kinetic energy at this point 290,400

A car slows from 37.3 m/s to 8.8 m/s with a constant acceleration of -3.3 m/s2. How long in seconds does it require?

Answers

Answer:

The time it takes the car to stop is 8.64 s.

Explanation:

Given;

initial velocity of the car, u = 37.3 m/s

final velocity of the car, v = 8.8 m/s

constant acceleration of the car, a = -3.3 m/s²

The time it takes the car to stop is given by;

[tex]t = \frac{v-u}{a}[/tex]

Substitute the givens;

[tex]t = \frac{v-u}{a}\\\\t = \frac{8.8-37.3}{-3.3}\\\\t = 8.64 \ s[/tex]

Therefore, the time it takes the car to stop is 8.64 s.

answer plese lang po​

Answers

Answer:

hi

Explanation:

how to convert gumbauan baboy to kung burger p

When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?

Answers

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

The effective spring constant of the springs is 33755.55 N/m.

The frequency of the car's vibration is 2.07 Hz.

What is force?

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

Weight of the four people: F = 310 × 9.80 N = 3038 Newton.

The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.

Hence, the effective spring constant of the springs: k= force/compression

= 3038 N/0.90 × 10⁻² m

= 33755.55 N/m.

The frequency of the car's vibration is: f = 1/2π√(k/m)

=1/2π√(33755.55/2000)

= 2.07 Hz.

Learn more about force here:

https://brainly.com/question/13191643

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The atomic of nitrogen is 7. The number of electrons a neutral atom has is_, and its atomic mass is approximately_amu.

Answers

The atomic of nitrogen is 7. The number of electrons a neutral atom has (6) and its atomic mass is approximately (1) amu

answer plz answer plzzz I am a little confused with full time ​

Answers

I can’t read that I’m sorry make it more clear

The mass of a paper-clip is 0.50 g and the density of its material is 8.0g/cm'. The total volume of
a number of clips is 20 cm.
How many paper-clips are there?​

Answers

Answer:

320 paper clips

Explanation:

mass = volume × density = 20cm³ × 8g/cm³ = 160g

mass of 1 paper clip = 0.50g

mass of x paper clips = 160g

x = 160/0.50 = 320

PLEASE HELP PLS ITS DUE IN 10 MINUTES!!
Suppose you’re standing at the top of a 100m high building. How much potential energy would you lose if you walked down the stairs by 20m?
You are now standing at a height of 80m. How much potential energy would you lose if you went down by 20m?
Suppose now you decide to go down by 20m more (you’re standing at 60m high). How much potential energy would you need to get to 40? Can you draw any conclusions about potential energy from this observation?

Answers

Answer:

I AM SO SORRY

but here is the answer

Explanation:

20 j

A beam contains 4.9 × 108 doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of 4.6 × 105 m/s. What is (a) the magnitude of the current density and (b) direction of the current density (c) what additional quantity do you need to calculate the total current i in this ion beam?

Answers

Answer:

[tex]72.12\ \text{A/m}^2[/tex]

south

cross sectional area of the beam

Explanation:

v = Velocity of ions = [tex]4.6\times 10^5\ \text{m/s}[/tex]

Number of ions per [tex]\text{cm}^3[/tex] = [tex]4.9\times 10^8[/tex]

Charge density would be the product of number of ions per [tex]cm^3[/tex] and the charge of electrons multiplied by 2 as they are doubly charged.

[tex]\rho_q=4.9\times 10^8\times 10^6\times 2\times 1.6\times 10^{-19}\\\Rightarrow \rho_q=0.0001568\ \text{C/m}^3[/tex]

Current density is given by

[tex]J=\rho_qv\\\Rightarrow J=0.0001568\times 4.6\times 10^5\\\Rightarrow J=72.12\ \text{A/m}^2[/tex]

The current density is [tex]72.12\ \text{A/m}^2[/tex]

The direction of the current density is opposite to the movement of the charged particle. The particles are moving north so the direction of current density will be to the south.

Current is given by

[tex]I=JA[/tex]

where A is the cross sectional area of the beam .

So the cross sectional area of the beam is required to determine the total current in this ion beam.

A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible. What was the momentum of the shopping cart before the collision? a) 66.0 kg-m/s b) 0 kg.m/s c) 126 kg.m/s d) 378 kg.m/s What was the momentum of the box before the collision? a) 66.0 kg.m/s b) 0 kg.m/s c) 378 kg-m/s d) 126 kg-m/s What is the velocity of the combined shopping cart-box wreckage after the collision? a) 6.0 m/s b) 3.9 m/s c) 0 m/s d) 11.5 m/s

Answers

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

a)

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

b)

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

c)

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

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If two balloons are charged and brought close to each other while hanging you observe that they move away from each other. What would the observation be if the balloons have larger charge?
A. There is no further effect. The repulsion magnitude is always the same.
B. There is no further effect. The magnitude of the force is determined by the charge signs not their magnitude
C. The balloons will lift in addition to separating as now they can start to overcome gravitational forces
D. The balloons will separate further as the repulsion magnitude increases
E. The balloons will come closer as the charges create larger polarization forces
F. The balloons will spin around each other since the electric force can produce rotational motion

Answers

Answer:

D. The balloons will separate further as the repulsion magnitude increases.

Which statement is true about two isotopes of the same element?

Answers

Answer:

atoms that have the same # of protons but a DIFFERENT # of NEUTRONS

Explanation:

Answer:

D-They have different number of neutrons

Explanation:

brainliest? Plz

A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.

Answers

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Magnitude of normal force ( object at rest );  n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x  after object start moving   x = 25 N

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

Learn more : https://brainly.com/question/21444366

What is the difference between renewable energy non-renewable energy? Use in your own words.

Answers

Non renewable energy are limited energy which would go away soon. Ex. Coal, oil, fossil fuels

Renewable energy comes from natural resources which would be there for a much longer time. Ex. Wind, sun, and water

A hockey player skating at 21 m/s comes to a complete stop in 12.0 m.

a) What is its acceleration during this displacement?

b) How long did it take the hockey player to come to a stop?

Answers

Answer:

a. 18.375m/s²

b. 1.142s

Explanation:

A. Using v² = u² + 2as

Where;

V = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration (m/s²)

s = distance (m)

In this question, v = 21m/s, u= Om/s, s = 12m

21² = 0² + 2 × a × 12

441 = 0 + 24a

441 = 24a

a = 18.375m/s²

B). Using V = u + at

21 = 0 + 18.375t

21 = 18.375t

t = 21/18.375

t = 1.142s

Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity

Answers

Answer: • They have same change in velocity but different changes in kinetic energy

•They started at the same height.

Explanation:

First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.

Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.

Therefore, based in the explanation, the correct options are:

• They have same change in velocity but different changes in kinetic energy

•They started at the same height.

what body parts were scientists wanting to image that prompted the development of the CT scanner

Answers

Answer:

The head

Explanation:

A ball is thrown 24 m/s into the air. How high does it go?
556.4 m
0 m
29.4 m
-556.4 m

Answers

Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct

a piping system consists of 100 ft of 2-inch pipe, a sudden expansion to 3-inch pipe, and then 50 ft of 3-inch pipe. Water is flowing at 100 gal/min through the system. What is the pressure difference from one end of the pipe to the other

Answers

Answer:

16+15+19= ??

Am just messign with u lol

Explanation:

anwser s 19 inches

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