If you change the motor in your vehicle you need to notify the DMV within ____,
days of this change.

-20
-25
-10
-15

Answer:

Look at explanation

Explanation:

a) Kinetic Friction= μmg

μmg=0.53*88.9*9.8=461.75N

b)  -461.75N=ma

a= -5.19m/s^2

v=v0+at

5.19*1.7=v0

v0=8.81m/s^2

(a) The magnitude of the frictional force will be 461.75N

(b)The initial velocity will be 8.81 m/s.

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction= 0.53.

m is the mass = 88.9 kg

g is the acceleration due to gravity= 9.81 m/s²

v is the speed =?

The formula for friction force is;

[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]

Mechanical force is found as;

F=ma

-461.75=(88.9)a

(-ve shows the -ve work done)

a=-5.19 m/s

From the Newton's first equation of motion;

v=u+at

0=u+at

u=-at

u=(- (-5.19)(1.7)

u=8.81 m/s²

To learn more about the kinetic friction refer to;

https://brainly.com/question/13754413

#SPJ2

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic

Answer:

Option (c)

Explanation:

1.The transformer core is assembled from steel sheets to:

(a) Reduced power loss due to eddy current

(b) Reduced power loss due to hysteresis

(c) Reduced power loss due to current flowing through the winding

(d) Reduce all types of power loss.

A transformer is a device which converts the low voltage into high and vice  versa.

There are two types of a transformer.

Step up: It is used to convert low voltage into high.

Step down It is used to convert high voltage into high.

It depends on the number of turns in primary and the secondary coil.

The core of the transformer is laminated and it is in the form of sheets.

By using such type of core, the power loss due to the windings is reduced.

option (c) .

Answer:

Its the Federal government

Explanation:

a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy

b) burning of coal converts chemical energy to heat energy

c) rubbing of both hands against each other converts mechanical to heat energy

Answer:

a. solar cells

b.coal,wood,petroleum

c.rubbing ours palms

Complete question:

A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answer:

The tension on the rope is 1.95 N

Explanation:

The general equation of a progressive wave is given as;

[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]

Compare the given equation with the general equation of wave, the following parameters will be deduced.

A = 0.75 cm

k = 0.400π cm⁻¹

ω = 250π s⁻¹

The frequency of the wave is calculated as;

ω = 2πf

2πf = 250π

2f = 250

f = 250/2

f = 125 Hz

The wavelength of the wave is calculated as;

[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]

The velocity of the wave is calculated as;

v = fλ

v = 125 x 0.05

v = 6.25 m/s

The tension on the rope is calculated as;

[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]

Therefore, the tension on the rope is 1.95 N

Energy = (power) x (time)

Energy = (0.69 W) x (0.36 sec)

Energy = 0.25 Joule

Answer:

stress = 98000 N/m^2

strain = 3.92 x 10^-6

change in height = 0.0196 mm

Explanation:

Height, h = 5 m

Area, A = 3 m²

mass, m = 30000 kg

Stress is defined as the force per  unit area.

[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]

Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2

Young's modulus is defined as the ratio of stress to the strain.

[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]

let the change in height is h'.

Strain is defined as the ratio of change in height to the original height.

[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]

Answer:

you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

Answer:

If the mass of B is m and the temperature change is the same, the mass of B will be 2m.

Explanation:

Q = mcT

T = mc/Q

M = 4Q/2cT........... (1)

T = Q/mc

Plug this in equation 1.

M = 4Q/(2c × Q/mc)  = 4Q ÷ 2Q/m  = 4Q × m/2Q = 2m

Answer:

θ = 57.4°

Explanation:

The complete formula to find out the work done by the plane is as follows:

[tex]W = FdCos\theta[/tex]

where,

W = Work = 200000 J

F = Force = Tension = 2560 N

d = distance = 145 m

θ = angle between rope and horizontal = ?

Therefore,

[tex]200000\ J = (2560\ N)(145\ m)Cos\theta\\\\Cos\theta = \frac{200000\ J}{371200\ J}\\\\\theta = Cos^{-1}(0.539)[/tex]

θ = 57.4°

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

Answer:

0.774m

Explanation:

The formula for maximum height is given by:

hmax = ∨₀² ₓ Sin (α)² / 2 × g

where;

∨₀ = initial velocity

Sin (α) = angle of launch

g = gravitational acceleration which is equal to 9.8m/s²

Plugging in our values, we will have:

hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²

hmax= 20.25m/s × 0.75 / 19.8m/s²

hmax = 15.1875 / 19.8

hmax = 0.774m

Answer:

[tex]\lambda=6.83\times 10^{-5}\ m[/tex]

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]

So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].

Answer: I think its A or C I'm not sure though sorry.

Explanation:

Let T is the period of a pendulum. The SI unit of time is seconds (s).

It depends on the acceleration of gravity, g, and the length of the pendulum, l.

The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.

If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.

So,

[tex]T\propto \sqrt{\dfrac{l}{g}}[/tex]

Hence, this is the required solution.

Answer:

Answer:

because it look more impressive than empty dark sky .

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

[tex]A_a(t) = 5m/s^2[/tex]

To get the velocity, we integrate over time:

[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

[tex]V_a(t) = (5m/s^2)*t[/tex]

To get the position equation we integrate again over time:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

[tex]V_b(t) =20m/s[/tex]

To get the position equation, we can integrate:

[tex]P_b(t) = (20m/s)*t + P_0[/tex]

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

[tex]P_b(t) = (20m/s)*t + 100m[/tex]

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

[tex]P_a(t) = P_b(t)[/tex]

We can solve this for t.

[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]

We only care for the positive solution, which is:

[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

Distance travelled by a body in unit time is called speed. it is a scalar quantity because it can be specified only by magnitude.

Answer:

For example, the pressure acting on a dam at the bottom of a reservoir is ... pressure = height of column × density of the liquid × gravitational field ... The density of water is 1,000 kg/m 3.

Answer:

act or process of measuring

Explanation:

Explanation:

the comparison of an unknown quantity with a known quantity.

Answer:

The answer to this question is plasma

Answer:

Plasma

Explanation:

Answer:

They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.

Explanation:

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

Answer:

I could not find the answer or do it myself if I did find it I would defenetly share

Answer:

The expected radius of the Earth is 3.883 meters.

Explanation:

The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:

[tex]U = K[/tex] (1)

Where:

[tex]U[/tex] - Gravitational potential energy, in joules.

[tex]K[/tex] - Translational kinetic energy, in joules.

Then, we expand the formula by definitions of potential and kinetic energy:

[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

Where:

[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.

[tex]M[/tex] - Mass of the Earth. in kilograms.

[tex]m[/tex] - Mass of the rocket, in kilograms.

[tex]r[/tex] - Radius of the Earth, in meters.

[tex]v[/tex] - Escape velocity, in meters per second.

Then, we derive an expression for the escape velocity by clearing it within (2):

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)

If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]

[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]

[tex]r = 3.883\,m[/tex]

The expected radius of the Earth is 3.883 meters.

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI  

Answer:

hope it helps

much as you can