What two methods are the best choices to factor this expression 18x2-8

Answers

Answer 1

Answer:

Please check the explanation

Explanation:

The best two methods will be:

Factor by groupingFactor out the GCF

Factor by grouping

Factor by grouping deals with establishing a smaller groups from each term.

[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)[/tex]

[tex]8\:=\:\:2\cdot 2\cdot 2[/tex]

Therefore, the expression becomes

[tex]18x^2=\:\left(2\cdot 3\cdot 3\cdot x^2\right)-\left(2\cdot \:2\cdot \:2\right)[/tex]

Now factor out the greatest common factor (GCF) which is 2

         [tex]=\:2\left(3\cdot \:\:3x^2-\left(2\right)\left(2\right)\right)[/tex]

           [tex]=2\left(9x^2-2\cdot \:2\right)[/tex]

            [tex]=2\left(9x^2-4\right)[/tex]

Factor out the GCF

Given the expression

[tex]18x^2-8\:\:\:[/tex]

factor out common term 2

[tex]=2\left(9x^2-4\right)[/tex]

[tex]=2\left(3x+2\right)\left(3x-2\right)[/tex]          ∵ [tex]Factors\:\:\left(9x^2-4\right)=\left(3x+2\right)\left(3x-2\right)[/tex]


Related Questions

Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, what direction and value is his acceleration air resistance is negligible.

A. 9.8 m/s/s west
b. 9.8m/s/s east
C. 9.8m/s's down
d 9.8m/s/s south​

Answers

Answer:

9.8m/s^2 down  (option C)

Explanation:

The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.

2. An ambulance traveling at 20 m/s emits a sound at 500 Hz. What frequency does a person standing on the corner of a street detect?

Answers

531 Hz. As ambulance approaches and 470ish as it proceeds past stationary person.

Doppler effect.

Used online calculator. Use negative 20 m/s approach and positive 20 m/s leaving or having passed the person

that delivers oxygen to your body and In the video your blood is compared to a picks up CO2 to be released out when you breath. PLEASE I NEED A ANSWER ​

Answers

I’m a bit confused. What’s the question?

We will now determine the indexes of refraction for two Mystery materials, A and B. These materials can be selected from the list of materials on the right. Be sure to set your laser pointer to a frequency of 589 nm. Questions:A. Devise an experiment for determining the indices of refraction for these. Explain your methodology. B. What are the indices of refraction for the two mystery materials, A and B?

Answers

Answer:

A) refraction experiment   n = n₁ sin θ₁ / sin θ₂

B)  n_A = 1.19 ,    n_B = 1.53

Explanation:

A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index

            n₁ sin θ₁ = n₂ sinθ₂

            n₂ = n₁ sin θ₁₁ /sin θ₂

If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is

            n = n₁ sin θ₁ / sin θ₂

B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be

         

material A

             n_A = sin 50 / sin 40

             n_A = 1.19

material B

              n_B = sin 50 / sin30

              n_B = 1.53

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

Answers

Answer:

a

[tex]\lambda = 3.68 *10^{-36} \ m[/tex]

b

[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Explanation:

From the question we are told that

   The mass of the person is  [tex]m = 180 \ kg[/tex]

    The speed of the person is  [tex]v = 1 \ m/s[/tex]

    The energy of the proton is  [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda = \frac{h}{m * v }[/tex]

Here  h is the Planck constant with the value

      [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So  

     [tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]

=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]

Generally the energy of the proton is mathematically represented as

         [tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]

Here [tex]m_p[/tex]  is the mass of proton with value  [tex]m_p = 1.67 *10^{-27} \ kg[/tex]

=>     [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]

=>   [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]

=>   [tex]v = 3.09529 *10^{7} \ m/s[/tex]

So

        [tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]

so    [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]

=>     [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

     

A train is traveling at 55m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed of9 m/s and it takes 49 s to properly slow down, what distance does the train travel while slowing down?

Answers

Answer:

x = 1127 [m]

Explanation:

In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.

[tex]v_{f} =v_{o} -a*t[/tex]

where:

Vf = final velocity = 9 [m/s]

Vo = initial velocity = 55 [m/s]

a = acceleration o desacceleration [m/s²]

t = time = 49 [s]

Now replacing:

9 = 55 - a*49

a*49 = 55 + 9

a = 1.306 [m/s²]

Note: The negative sign in the above equation means that the speed decreases.

Now using the second equation.

[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]

(9)² = (55)² - 2*(1.306)*x

2944 = 2.612*x

x = 1127 [m]

As a rough model of the impact of walking/running, consider that half the mass of the body falls from a height of 4.77-cm onto a single foot. (During a typical stride, an adult's center-of-mass moves approximately this distance vertically). Use the kinematic equations to calculate the speed of an object falling from this height at the moment of impact with the ground under the influence of gravity.A. As a rough model of the impact of walking, consider that half of the mass of the entire body strikes the ground with a downward velocity of 1.0 m/s and comes to a full vertical stop over an impact duration of 20 ms. Calculate the force associated with this single step for a person with a mass of 74.2 kg. B. Calculate the stress (solid pressure) of a force of 1880 N applied across the 0.4 cm^2 cross-sectional area of the typical Achilles tendon. For reference, the maximum rupture stress of tendons has been reported in the range of 100-150 MPa.

Answers

Answer:

0.967 m/s

1855 N

[tex]46.375\ \text{MPa}[/tex]

Explanation:

v = Final velocity

u = Initial velocity = 0

s = Displacement = 4.77 cm

g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

From the kinematic equations

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]

The velocity of the object at the moment of impact is 0.967 m/s

Now

[tex]\Delta v[/tex] = Change in velocity = 1 m/s

t = Time taken = 20 ms

m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]

[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]

The force associated with a single step of the person is 1855 N

A = Area = [tex]0.4\ \text{cm}^2[/tex]

Stress is given by

[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]

The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]

The speed of object during falling is 0.967 m/s.

(A)  The magnitude of force  associated with this single step for a person is 1855 N.

(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

Given data:

The height of fall is, h = 4.77 cm = 0.0477 m.

The magnitude of downward velocity is, v' = 1.0 m/s.

The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].

The mass of person is, m = 74.2 kg.

The magnitude of force is, F' = 1880 N.

The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].

The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,

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

Solving as,

[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]

Thus, the speed of object during falling is 0.967 m/s.

(A)

Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.

Apply the expression of average force as,

[tex]F =\dfrac{m'v'}{t}[/tex]

Solving as,

[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]

Thus, the magnitude of force  associated with this single step for a person is 1855 N.

(B)

The expression for the stress is given as,

[tex]\sigma = \dfrac{F'}{A}[/tex]

Solving as,

[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]

Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].

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A penny is dropped from rest from a building 100m tall. what kind of motion is this

A. centripetal
B. Free fall
C. Linear
D. projectile​

Answers

Answer:

this is a projectile

Answer:

D. projectile

Explanation:

Since it's dropped from a rest that means that it's velocity at the beginning is 0.

(1.5 pts) A woman pushes on a box to the left. If the box is accelerating, what forces are working on the
Question 2:
box? (Draw both y and x forces)

Answers

Answer:

Nope

Explanation:

A television of mass 15 kg sits on a table. The coefficient of static friction
between the table and the television is 0.35. What is the minimum applied
force that will cause the television to slide?

A) 38 N
B) 147 N
C) 51 N
D) 79 N

Answers

Answer:

more than 51.45 N

__________________________________________________________

We are given:

Mass of the television = 15 kg

Coefficient of Static friction = 0.35

Minimum force required to move the television:

Normal Force:

We know that the normal force is equal and opposite to the Weight of the television

Weight of the television = Mg            

[where m is the mass and g is the acceleration due to gravity]

Weight = 15 * 9.8

Weight = 147 N

Force of Friction:

We are given the coefficient of Friction = 0.35

We know that coefficient of Friction = Force of friction / Normal Force

replacing the variables

0.35 = Force of Friction / 147

Force of Friction = 147 * 0.35                     [multiplying both sides by 147]

Force of Friction = 51.45 N

Since a force of 51.45 N is will be applied opposite to the direction of application of Force, the television will only move when we apply more force than 51.45 N

Answer:

it is C

Explanation:

The particle accelerator at CERN can accelerate an electron through a potential

difference of 80 kilovolts. Calculate

(a) The kinetic energy (in keV) of the electron​

Answers

Answer:

K.E = 1.28 × 10^-17 KeV

Explanation:

Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.

To Calculate the kinetic energy (in keV) of the electron​, let us first find the electron charge which is 1.60 × 10^-19C

The kinetic energy = work done

K.E = e × kV

Substitute e and the voltage into the formula

K.E = 1.60 × 10^-19 × 80

K.E = 1.28 × 10^-17 KeV

Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV

Which is increased when the string of a stringed instrument is tightened?
timbre
pitch
wavelength
loudness

Answers

the answer is b: pitch

When the string of the instrument is tightened then the length of the string decreases hence the pitch and frequency will increase so, option B is correct.

What is pitch?

The frequency at which the sound waves that create a sound vibrate determines its pitch. High-frequency sound waves produce high-pitched noises, and low-frequency sound waves make low-pitched noises. The ability to discern between harsh and flat sounds is known as pitch.

A string will vibrate at a varied frequency depending on its length. Higher frequency and higher pitch are produced by shorter strings.

The pitch rises as the anxiety does as well. A string's length is also crucial. A string vibrates and makes music when it is supported at two points and pulled. The pitch of this string will, however, rise if the length is shortened.

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A T-shirt cannon mounted at the top of an arena needs to fire a t-shirt into the first row, a horizontal distance of 39 meters away. If the cannon launches t-shirts at 12 m/s, how high is the cannon mounted?
Question 1 options:
3.3 m
16.2 m
53.4 m
8.9 m

Answers

Answer:

h = 51.75 m

nearest answer is:

53.4 m

Explanation:

First we analyze the horizontal motion. Since, the air friction is assumed to be negligible. Hence, the horizontal motion shall be uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance = 39 m

V₀ₓ = Horizontal Initial Velocity = 12 m/s

t = time = ?

Therefore,

39 m = (12 m/s)t

t = 39 m/12 m/s

t = 3.25 s

Now, we analyze the vertical motion. Applying newton's second equation of motion to vertical motion:

h = V₀y t + (1/2)gt²

where,

h = height of cannon = ?

V₀y = initial vertical velocity = 0 m/s

g = 9.8 m/s²

Therefore,

h = (0 m/s)(3.25 s) + (1/2)(9.8 m/s²)(3.25 s)²

h = 51.75 m

Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled eastward at 8 m/s while billiard ball B originally traveled westward at 2 m/s. Calculate the speed and direction of each ball after the collision.

Answers

Answer:

lucky mauld mauldgomary was an british poet...

Which image illustrates the interaction of a light wave with a mirror?
t J
A
с
.
A. A
B. B
C. C
D. D
0

Answers

Answer:

I'm pretty sure its A

Explanation:

because its a reflection- Hope you get a good grade!

7) A moving object is rolling on a surface that is 5 m off the ground. The object is moving at a constant speed of 4 m/s. If the object is 3.2 kg, what is the final energy of the ball after rolling for 10 m, assuming friction is negligible?
156.8 J
131.2 J
182.4 J
25.6 J

8) A spreadsheet application is used to create a computational model of the energy experienced by a pendulum. How do the energy values of the pendulum relate?

The sum of the potential energy and the kinetic energy is always constant.

The sum of the potential energy and the kinetic energy is always 0.

The potential energy is always greater than the kinetic energy.

The kinetic energy is always equal to the potential energy.

Answers

131.2 J and The last one on number 8

I gave the same answer and it passed.

7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.

8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.

What is energy?

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Given:

A moving object is rolling on a surface that is 5 m off the ground,

The speed of the object, v = 4 m/s,

The mass of the object, m = 3.2 kg,

Calculate the kinetic energy after 10 meters as shown below,

KE = 1/2 × 4² × 3.2

KE = 25.6 J

Calculate the potential energy as shown below,

PE = 3.2 × 9.8 × 5

PE = 156.8 J

Thus, total energy = KE + PE

The total energy = 25.6 + 156.8

The total energy = 182.4 J

8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.

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how does tom and jerry movie character influence your attitude​

Answers

Answer:

it makes me wish I was a cartoon

Answer:

goofy and stupid and act like a kid

Explanation:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answers

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

Dividing (4) by (3), from (2), we have:

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

Solving for aw, we get:

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is present, given by B = 4.0i + 9.0x2 j , with x in meters and B in mT. Calculate the k-component of the force on the 2 m segment of the conductor that lies between x = 1.0 m and x = 3.0 m.

Answers

Answer:

0.546 [tex]\hat k[/tex]

Explanation:

From the given information:

The force on a given current-carrying conductor is:

[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]

where the length usually in negative (x) direction can be computed as

[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]

[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]

[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]

where;

current I = 7.0 A

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]

F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]

F = 0.546 [tex]\hat k[/tex]

what does the modal "must"indicate?

Answers

Answer:

The modal verb must is used to express obligation and necessity. The phrase have to doesn't look like a modal verb, but it performs the same function.

A car traveling initially at a speed of 20 m/s accelerates to a speed of 31 m/s over a distance of 45 meters.
What is the magnitude of the car's acceleration?

Answers

Answer:abc defg hijk lmnop qrs tuv wx y and z

Explanation: now i know my abc's

an object of mass 4kg moving with initial velocity if 20m/s accelerates for 10s and attaind a final velocity of 60m/s calculate the acceleration​

Answers

Answer:

given us,

mass= 4×9.8gm m(9.8) formula

= 39.2

final velocity (v)= 60m/s

initial velocity (u)= 20m/s

time(t)= 10s

acleration(a)=?

now,

accleration(a)= v-u/t=60- 20/10

=40/10

=4m/s

:. the acceleration is 4 m/s

Explanation:

first we have to calculate mass and we can use acceleration formula

car driving on a circular test track shows a constant speedometer reading of 100 kph for one lap. a. Describe the car's speed during this time. b.

Answers

Answer:

Speed = 100 km/h

Explanation:

Given:

Speedometer reading = 100 kph for one lap

Assume;

Time taken to complete one lap = 1 hour

Computation:

Speed =  Distance / Time

Speed =  100 / 1

Speed = 100 km/h

A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?

Answers

Answer:

(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰

(a)(ii) The maximum torque is 0.132 Nm

(b) The orientation of the coil is 45⁰

Explanation:

Given;

diameter of the circular wire, d = 8.6 cm = 0.086 m

radius of the wire, r = d /2 = 0.043 m

number of turns, N = 15 turns

magnetic field, B = 0.56 T

The torque on the wire is given by;

τ = NIABsinθ

where;

θ is the orientation of the wire

(a) maximum torque occurs when the orientation of the wire is at 90⁰

The maximum torque is given by;

τ = NIABsin(90⁰)

τ = NIAB

τ = (15)(2.7)(π x 0.043²)(0.56)

τ = 0.132 Nm

(b)

71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm

[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]

A sample of an ideal gas has a volume of 0.0100 m^3, a pressure of 100 x 10^3 Pa, and a temperature of 300K. What is the number of moles in the sample of gas?

Answers

Answer:

Explanation:

pV = nrT

n = PV/RT

n = (100*10^3)(.01)/(300*0.082057)

n = 40.62 moles

An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?

Answers

Answer:

2.22 kPa

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

From the question we have

[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]

We have the final answer as

2.22 kPa

Hope this helps you

please help! What is the relationship between velocity and acceleration?

Answers

Answer:Acceleration implies any change in the velocity of the object with respect to time. Velocity is nothing but the rate of change of displacement. On the other hand, acceleration is the rate of change of velocity with respect to time.

Explanation:

A rod is pivoted about its center and oriented horizontally. A 5.0-N force directed upward is applied 4.0 m to the left of the pivot and another upward 5.0-N force is applied 1.5 m to the right of the pivot. What is magnitude of the total torque about the pivot?

Answers

Answer:

The total torque is 27.5 Nm

Explanation:

Given;

5.0-N force directed upward is applied 4.0 m to the left of the pivot,

5.0-N force directed upward is applied 1.5 m to the right of the pivot,

Taking the moment about the pivot, the total torque is given by;

τ = Fr

where;

F is the appllied force

r is the radius of the force arm

τ = (5 N x 4 m) + (5 N x 1.5 m)

τ = 27.5 Nm

Therefore, the total torque is 27.5 Nm

Which element is used in the manufacture of mirrors and bronze?

Answers

Answer:

Silver

Explanation:

Silver is an important element in the manufacturing process of mirrors. Silver is used to make mirrors through the process we call "silvering". Silvering is a process in which a glass is coated with reflective substances so as to produce reflections, and then mirrors.

In Silvering process, Chlorine is also used. Stannous Chloride is the particular compound used to carry out the silvering, it has the chemical formula, SnCl₂

A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?

A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s

Answers

Answer:

B. 1275 kg*m/s

Explanation:

I = F(deltaT) = (deltaP) = mv2- mv1

Therefore,

I = mv2-mv1

m = 850 kg

v2 = 5 m/s

v1 = 3.5 m/s

I = (850)(5)-(850)(3.5)

I = 1275 kg* m/s

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