An ant is crawling along a yardstick that is pointed with the O-inch mark to the east and the 36-Inch mark to the west. It starts at the 19-inch mark, crawls to the 27-inch

mark, then moves to the 7-inch mark. What was the total distance the ant traveled?

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:

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

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

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.

Answer:

The head

Explanation:

Answer:

C. Equals

Explanation:

Law of reflection Equals the angle of incidence

Answer:

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

Hope this helps you

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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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

Answer:

B is the best answer for this

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

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

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

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

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

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.

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

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

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.

Please vote brainliest

Answer:

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

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.

Answer:

v = 0.4 m/s

Explanation:

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

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

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

Answer:

16+15+19= ??

Am just messign with u lol

Explanation:

anwser s 19 inches

i

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]

Answer:

[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]

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.

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:

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Answer:

A

Explanation: