Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures

Answers

Answer 1

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m


Related Questions

A heavy truck moving with 20 km/hr hits a car at rest. A physics student argued that
the maximum velocity the car suddenly gains is 40 km/hr. Do you agree with it?
Explain with necessary theory

Answers

Answer:

Yes

Explanation:

speed of truck = 20 km/h

Initially the car at rest.

maximum velocity of car = 40 km/h

When the truck and the car collide, the momentum of the truck transferred to car.

So, the car can attain the speed of 40 km/h.

A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km

Answers

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

Please assist with solving this problem and showing the steps

Answers

Answer:

2.21 N

Explanation:

The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.

The fulcrum feels F1 + F2 + 34 * 980

F2 = 141.7 * 980 = 138866

F1 = 50.3 * 980  =  49294

Ruler = 34 * 980=  33320

Total Force = 221480 The units here are dynes

I just saw in the middle of the question that g = 9.80

So the answer becomes 221480 / 1000 = 221.48   because we needed kg

And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980

The total force exerted on the fulcrum is

convert 2.4 milimetre into metre​

Answers

Answer is 0.0024

Explanation

divide the length value by 1000.

A 1.0 ball moving at 2.0 / perpendicular to a wall rebounds from the wall at 1.5 /. If the ball was in contact with the wall for 0.1 , what force did the wall impart onto the ball?

Answers

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

The amount of force that the wall imparts on the ball is 5.0N

According to Newton's second law, the formula for calculating the force applied is expressed as:

[tex]F=ma[/tex]

m is the mass of the object

a is the acceleration of the object

Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]

The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]

Given the following parameters

m = 1.0kg

[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]

t = 0.1sec

Substitute the given parameter into the formula

[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]

Hence the amount of force that the wall imparts on the ball is 5.0N

Learn more here: https://brainly.com/question/17811936

Typhoon signal number 2 is raised. What is the speed of the expected typhoon?​

Answers

the simple answer is from 61kmph to 120kmph

Explanation:

no explanation is needed


Question 8 a-e plz

Answers

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces an amount of electric energy with the hot reservoir at 500 K during Day One and then produces the same amount of electric energy with the hot reservoir at 600 K during Day Two. The thermal pollution was:

Answers

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J

An ice skater with a mass of 50 kg is gliding acrossthe ice at a speed of 8 m/s when herfriend comes up from behind and gives her a push,causing her speed to increase to 12m/s. How much work did the friend do on the skater

Answers

Answer:

[tex]W=2KJ[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=50kg[/tex]

Initial Velocity [tex]v_1=8m/s[/tex]

Final Velocity [tex]v_2=12m/s[/tex]

Generally the equation for Work-done is mathematically given by

W=\triangle K.E

Therefore

 [tex]W=0.5M(v_2^2-v_1^2)[/tex]

 [tex]W=0.5*50(12^2-8^2)[/tex]

 [tex]W=2KJ[/tex]

Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about 25 m. if the maximum advisable acceleration of your vehicle through a turn on wet pavement is 0.40 times the free-fall acceleration, what is the maximum speed at which you should drive through this turn?

Answers

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s[/tex]

So, the maximum speed of the car should be 9.89 m/s.

Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:

a. 1,5 N

b. 15,0 N

c. 3,1 N

d. 5,2 N

Answers

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]

Option (A) is correct.

The unit of kinetic energy is the _______. The unit of kinetic energy is the _______. hertz meter watt joule radian

Answers

Answer:

joule

Explanation:

two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same

Answers

Answer:

  r ’= 4 r

Explanation:

Electric potential energy is

          U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12

in this exercise

          q₁ = q₂ = q

          U = k q² / r

for when the charge change

           U ’= k q’² / r’

indicate that

      q ’= 2q

      U ’= U

we substitute

           U = k (2q) ² / r ’

           U = 4 k q² / r ’

we substitute

           [tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’

           r ’= 4 r

93 cm3 liquid has a mass of 77 g. When calculating its density what is the appropriate number of significant figures

Answers

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to

Answers

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

x₁ = -0.20 mx₂ = 0 mk = 60 N/m

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

Question 7 of 10
A railroad freight car with a mass of 32,000 kg is moving at 2.0 m/s when it
runs into an at-rest freight car with a mass of 28,000 kg. The cars lock
together. What is their final velocity?
A.1.1 m/s
B. 2.2 m/s
C. 60,000 kg•m/s
D. 0.5 m/s

Answers

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 40.0 mph and half the distance at 60.0 mph . On her return trip, she drives half the time at 40.0 mph and half the time at 60.0 mph.

Required:
a. What is Julie's average speed on the way to grandmother's house?
b. What is her average speed in the return trip?

Answers

Answer:

a. The average speed on her way to Grandmother's house is 48.08 mph

b. The average speed in the return trip is 50 mph.

Explanation:

The average speed (S) can be calculated as follows:

[tex] S = \frac{D}{T} [/tex]

Where:

D: is the total distance

T: is the total time

a. To find the total distance in her way to Grandmother's house, we need to find the total time:

[tex]T_{i} = t_{1_{i}} + t_{2_{i}} = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}}[/tex]

Where v is for velocity

[tex] T = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}} = \frac{(100/2) mi}{40.0 mph} + \frac{(100/2) mi}{60.0 mph} = 1.25 h + 0.83 h = 2.08 [/tex]    

Hence, the average speed on her way to Grandmother's house is:

[tex]S_{i} = \frac{D}{T_{i}} = \frac{100 mi}{2.08 h} = 48.08 mph[/tex]

b. Now, to calculate the average speed of the return trip we need to calculate the total time:                        

[tex]D = v_{1_{f}}\frac{T_{f}}{2} + v_{2_{f}}\frac{T_{f}}{2} = \frac{T_{f}}{2}(v_{1_{f}} + v_{2_{f}})[/tex]

[tex]100 mi = \frac{T_{f}}{2}(40 mph + 60 mph)[/tex]

[tex] T_{f} = \frac{200 mi}{40 mph + 60 mph} = 2 h [/tex]

Therefore, the average speed of the return trip is:

[tex]S_{f} = \frac{D}{T_{f}} = \frac{100 mi}{2 h} = 50 mph[/tex]

I hope it helps you!                                                      

A wheel accelerates so that it's angular speed increases uniformly from 150 rads/s to 580 rads/s in 16 revolutions.Cakcjlate its angular acceleration. ​

Answers

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

A is the angular acceleration.t is the timeVw is the final angular speed.Uw is the initial angular speed.

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

Which number has four significant figures?

A. 4000
B. 3.008
C. 86.012
D. 0.0001

Answers

a. 4000

This has 4-digits.

Answer:

in my opinion letter d.

Explanation:

Sana pi tama

What is the rate of the entropy change of the universe as heat leaks out a window, consisting of a single pane of glass that is 0.5 cm thick and 1.0 m2 in area, where the indoor temperature is 25°C and the outdoor temperature is -10°C?

Answers

Answer:

The change in entropy is 1.6 W/K.

Explanation:

Thickness, d = 0.5 cm

Area, A = 1 m^2

T = 25°C

T' = - 10°C

Coefficient of thermal conductivity of glass, K = 0.8 W/mK

The change in entropy is given by

S = Q/T

Here,

[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]


A current of 5.50 A flows in a conductor for 7.5 s. How much charge passes a given point in the conductor during this time?

Answers

56.1 neither songs were

The equations for calculating both the electric force and the gravitational force are above. Their equations are very similar. What is an important difference between these two forces?

A The electrical force is measured in coulombs; the gravitational force is measured in newtons.

B The electrical force between two charged objects will always be weaker than the gravitational force between them.

C The gravitational force decreases with the square of the distance between the objects; the electrical force increases with the square of the distance between the objects.

D Electrical forces can be attractions or repulsions; gravitational forces can only be attractions.

Answers

A, B, and C are hilarious. D is correct.

Charges can be positive or negative, so a pair of charges can be alike or opposite. But so far, we've never seen a negative mass.

if a body covers 100m in 5 second from rest find the acceleration produced by a body in 10 second​

Answers

Answer:

a=10m/s^2

Explanation:

acceleration= final velocity+ initial velocity/time taken

v-u/t=a

100-0/5=a

100/5=a

a=20m/s^2

case2

100-0/10=a

100/10=a

a=10m/s^2

Don't forget to write the units.

Hope this helps

please mark me as brainliest.

Of the following, which have the highest frequency in the electromagnetic
spectrum?
A. Visible light
B. Infrared waves
C. Ultraviolet rays
D. X-rays

Answers

d. X rays ..........

Hi, so i have to find T1, can some1 help?

Answers

30.1 N

Explanation:

Given:

[tex]W_1 = 16\:\text{N}[/tex]

[tex]W_2 = 8\:\text{N}[/tex]

Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.

Forces involving W1:

[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]

[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]

Forces involving W2:

[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]

[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]

Substitute (2) into (3) and we get

[tex]T_1\sin 53 - W_1 = W_2[/tex]

Solving for [tex]T_1[/tex],

[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]

what is time taken by radio wave to go and return back from communication satellite to earth??​

Answers

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 88.9 kg and the coefficient of kinetic friction between the ground and the player is 0.53. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s)

Answers

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.

What is kinetic friction?

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

A student measure the length of a laboratory bench with a meter ruler. Which of the following values is the most approbriate way to record the result ? a.4.022m b.4.02m c.4.0m d.4m​

Answers

Answer:

Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)

Explanation:

A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops

Answers

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

a. What do you mean by chromatic aberration in lenses?

Answers

Chromatic aberration is a phenomenon in which light rays passing through a lens focus at different points, depending on their wavelength. ... the same area of the photo after post-production removal of the chromatic aberration using a software tool.
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