If I travel 300 m east, then 400 m west, what is my distance &
displacement?

Answers

Answer 1

Answer:100m west

Explanation:


Related Questions

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

According to the textbook, which indicator should Andrea use? Hint: Neutral substances have a pH of 7. Acids have a pH that is less than 7, and bases have a pH that is greater than 7. OA, phenol Red OB thymol Blue OC. phenolphthalein D. trinitrobenzoic acid Raclet Submit​

Answers

Answer:

Thymol blue

Explanation:

Answer: c

Explanation:

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

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

Answers

Answer:

The head

Explanation:

Calculate how long
in years it will take your radiation to reach Earth from Alpha Centauri (research this).
( MY RADIATION IS INFARED )

Answers

Alpha Centauri is 4.4 lightyears away. Traveling at 1/10 speed of light it would take you 44 years.

A student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?

Answers

Answer:

[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]

[tex]y=10\ m/s[/tex]

Explanation:

Rectangular coordinates of vectors in 2D

Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:

[tex]x=v\cos\theta[/tex]

[tex]y=v\sin\theta[/tex]

The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.

The rectangular components of the vector are:

[tex]x=20\cos 30^\circ[/tex]

[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]

Operating:

[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]

[tex]y=20\sin 30^\circ[/tex]

[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]

Operating:

[tex]\mathbf{y=10\ m/s}[/tex]

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

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You just tested the hypothesis a proposed model for the position of bright fringes produced by two slits. Explain what the model says and what you found in your investigations with the simulation.

Answers

Answer:

two-slit interference model was proposed by Young     d sin θ = m λ

Explanation:

The two-slit interference model was proposed by Young, it establishes that if a coherent source of light passes through two slits, the shape of the given pattern is a consequence of the relative phase difference between the two rays; mathematically it can be expressed by

           d sin θ = m λ

          m= 0, 1, 2, 3, ...

for constructive interference, that is, the two rays arrive with a number between wavelengths.

D is the distance between the slits, tea the angle between the two rays, m an integer and m the wavelength used.

In a simulation a pattern of slits of equal intensity and equally spaced is observed.

An object moves along the x-axis with its position x given as a function of time t by x(t)= Ht^2 - Ft + G
What is the object's velocity

Answers

Answer:

v(t) = 2Ht - F

Explanation:

Since, the position of the object is given in terms of time (t) as follows:

x(t) = Ht² - Ft + G

where,

H, F, G are constants.

Therefore, the velocity of the object can also be found in terms of the time (t), by simply taking the derivative of the given position equation with respect to time. So, the velocity can be found as follows:

(d/dt) x(t) = (d/dt)(Ht² - Ft + G)

v(t) = (d/dt)(Ht²) - (d/dt)(Ft) + (d/dt)(G)

v(t) = H (d/dt)(t²) - F (d/dt)(t) + (d/dt)(G)

v(t) = H(2t) - F(1) + 0

v(t) = 2Ht - F

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

What is the value of the normal force of the coefficient of kinetic friction friction is 0.22 and kinetic friction force is 40 N

Answers

Answer: 1.8x10 sqaured newtons for all plato users

Explanation:

The value of the normal force of reaction is 181.8 N

Friction is defined as the resistance to motion caused by the contact between a surface and the object moving against it.

The normal reaction force is the weight of the body experiencing friction and normally acts vertically downwards.

The frictional force, the coefficient of kinetic friction and the normal force are related by the formula:

coefficient of friction = frictional force / normal reaction

Therefore, normal reaction = frictional force / coefficient of friction

Frictional force = 40 N;   coefficient of kinetic friction friction is 0.22

Normal reaction = 40 N/ 0.22 = 181.8 N

Therefore, the value of the normal force of reaction is 181.8 N

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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 plane starts from rest accelerates to 40 m/s in 10 seconds. How far did the plane travel during this time?

Answers

Answer:

200 m

Explanation:

We are given:

Initial velocity of the plane (u) = 0 m/s

Final velocity of the plane (v) = 40 m/s

Time interval (t) = 10 seconds

Displacement of the plane (s) = x m

Solving for x:

Acceleration of the plane

v = u + at                                                     [First equation of motion]

40 = 0 + a(10)                                              [replacing known variables]

a = 4 m/s²                                                    [dividing both sides by 10]

Displacement of the Plane:

s = ut + 1/2 (at²)                                            [Second equation of motion]

s = (0)(10) + 1/2(4)(10)²                                  [replacing known variables]

s = 200 m

Hence, the Plane covers a distance of 200 m in the given time interval

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]

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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

How does Hydroelectric Energy impact the environment? Use in your own words.

Answers

Explanation:

ok sooooooooooo Hydropower is fueled by water so it's a clean fuel source, meaning it won't pollute the air like power plants that burn fossil fuels.

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.

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In contact forces, _____.

A.) objects do not touch each other
B.) objects must touch each other
C.) more work is done than in other forces

Answers

Answer: B

Explanation:

Answer:

B is the best answer for this

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

i

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

At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1500 kg car traveling at 1.0 m/s?

Answers

Answer:

15m/s

Explanation:

Step one:

given data

mass of bicycle m=100kg

the velocity of bicyle v=?

mass of car M=1500kg

the velocity of car V=1.0m/s

Step two:

we know that the momentum is expressed as

P=mv

since the momentum of the bicycle must be equal to car then

mv=MV---------1

100*v=1500*1

divide both sides by 100

v=1500/100

v=15m/s

The velocity of the bicycle should be 15m/s

During takeoff, an airplane goes from 0 to 42 m/s is 5 s.

How far is it going after 4 s. Calculate answer to one decimal

Answers

Answer:

33.6 m

Explanation:

In order to solve the distance of the airplane traveled, we have to write each statement in the question:

1. The distance covered by the plane after 5s: 42 m ( This was calculated by getting the difference of the first and second speed recorded)(42 - 0 = 42)

2. Time taken for the plane to reach this speed: 5s

The question asks for:

Distance of the plane after 4s.

We then solve this by saying:

If 42 m are covered in 5s, how many are covered in 4s? or:

42 m = 5s

x m = 4s

We then cross multiply:

42 × 4 = 168 and 5 × (x) = 5x

After that we divide the "x" side on both sides:

168÷5 and 5x ÷ 5

= 33.6 m = x

Since its already rounded to one decimal place already, it remains unchanged.

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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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 horse trots away from its trainer in a straight line, moving away 8m away in 8.0s it then turns abruptly gallops halfway back in 2.0’s what is the magnitude of the horses average velocity for the entire trip

Answers

Answer:

v = 0.4 m/s

Explanation:

By definition, the average velocity is the rate of change of the position with respect to time, as follows:

       [tex]v_{avg} = \frac{x_{f}-x_{o}}{t_{f} -t_{o}} (1)[/tex]

Choosing arbitrarily x₀ = 0 and t₀ = 0, (1) reduces to:

       [tex]v_{avg} = \frac{x_{f}}{t_{f} }} (2)[/tex]

During the first part of the trip, the horse moves 8m away from its trainer, and during the the second part, gallops halfway back, which means that it finishes 4m away from its trainer.Since total time is 8.0 s + 2.0 s = 10s, replacing xf = 4m and tif = 10,0 s in (2) we get:

       [tex]v_{avg} =\frac{x_{f}}{t_{f} } = \frac{4.0m}{10s} = 0.4 m/s (3)[/tex]

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

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

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.

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:

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

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