If the temperature of the conductor is increased, the electrons’ speeds decrease

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Answer 1

Answer:

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Explanation:

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If The Temperature Of The Conductor Is Increased, The Electrons Speeds Decrease

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Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?

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His bicycle that operates on a system for transportation

assuming a filament in a 120W light bulb acts like a prefect blackbody, what is the temperature of the hottest portion of the filament if it has a surface area of 6.4×10^_5m^2. The stefan- boltzmann constant is 5.67×10^-8W/(m2.k2) A. 12OOk B. 2400K C. 2100K​

Answers

Answer:

T = 2398 K

Explanation:

To calculate the emission of the light bulb we use the law is Stefan

           P = σ A e T⁴

as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)

           T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]

    let's calculate

           T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]

           T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]

           T = 2,398 10³ K

           T = 2398 K

A disk of charge is placed in the x-y plane, centered at the origin. The electric field along the axis of a positive disk of charge... points towards the disk along the z-axis. points away from the disk along the z-axis. always points in the positive z-direction. none of these choices

Answers

Answer:

Points away from the disk along the z-axis.

Explanation:

Along the axis of the disk, which is the z - axis, the total vertical electric field components of the charged disk sum up while the horizontal components cancel out. Thus, leaving only vertical components of electric field along the axis of the disk.

Since the disk is positively charged and electric field lines point away from a positive charge, the electric field along the axis of a positive disk of charge points away from the disk along the z-axis.

You swing a bat and hit a heavy box with a force of 1273 N. The force the box exerts on the bat is Group of answer choices less than 1273 N if the box moves. exactly 1273 N whether or not the box moves. None of the above choices are correct. exactly 1273 N only if the box does not move. greater than 1273 N if the bat bounces back. greater than 1273 N if the box moves.

Answers

Answer:

exactly 1273 N whether or not the box moves.

Explanation:

In the case when the bat is swing and it is hitted to a heavy box having a force of 1273 N so here the force of the box that exert on the box should be accurately 1273 N even if the box is moved or not. As the third law of the newton should be equivalent & the opposite reaction

Therefore as per the given situation, the above represent the answer

An empty 12,954 kg railroad car, traveling at a speed of 28 m/s strikes a partially filled 17,616 kg railroad car moving in the same direction at a speed of 5 m/s. What is the total momentum of the two railroad cars AFTER the collision?

Answers

Answer:

450792 kgm/s

Explanation:

by conservation of momentum,

total momentum AFTER collision = total momentum BEFORE collision

                                                       =mv+m'v'

                                                       =12954×28+17616×5

                                                       =450792 kgm/s

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ​

Answers

Answer:

144 meters

Explanation:

it takes 6 seconds to hit the ground right and the ball lays off 24 m per second .

so by the time the ball hits the ground 6 seconds passed. so that means the cliff is 6.0×24=144

what is the frequency of a wave related to​

Answers

Answer:

Frequency is the number of complete oscillations or cycles or revolutions made in one second.

Which conclusion can be made based on the information in the table?
Wave speed and wavelengths can vary inversely to produce the same frequency.
O Frequency and wave speed can vary directly to produce the same wavelength.
O Wavelengths and frequency can vary inversely to produce the same wave speed.
O Frequency and wavelengths can vary directly to produce the same wave speed.
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Answer:

The correct option is (b).

Explanation:

The relation between the wavelength and frequency is given by :

[tex]\lambda=\dfrac{v}{f}[/tex]

Where

v is the wave speed

f is the frequency of a wave

It is clear from the above equation that the wavelengths and frequency can vary inversely to produce the same wave speed.

A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replaced with a 2.8kg mass.What is the stretch of the spring?

Answers

2.0kg and the mass of sometimes makes it hard 2.8kg

Please help I need this done within 30 mins

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It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.

A locomotive pulls 11 identical freight cars. The force between the locomotive and the first car is 150.0 kN, and the acceleration of the train is 2 m/s2. There is no friction to consider. 1) Find the force between the tenth and eleventh cars. (Express your answer to two significant figures.)

Answers

Answer:

The force between the 10 th car and the 11 th car is 13636.4 N.

Explanation:

Force, F = 150 kN

acceleration, a = 2 m/s^2

Let the mass of each car is m. \Total numbers of cars = 11

F = n m a

150000 = 11 x m x 2

m = 6818.18 kg

The force between the 10 th and 11 th car is

T = ma = 6818.18 x 2 = 13636.4 N

How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *

Answers

Answer:

you drove 50km

Explanation:

10×5 hope this helps

Answer:

50 Km

Explanation:

This is how far you have got on your journey if traveling like this.

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Hope this Helps

A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket

Answers

Answer:

The maximum height reached by the rocket is 486.53 m

Explanation:

Given;

initial velocity of the rocket, u = 0

acceleration of the rocket, a= 20 m/s²

duration of the rocket first motion, t = 4 s

The distance traveled by the rocket before its thrust failed

h₁ = ut + ¹/₂at²

h₁ = 0 + ¹/₂ x 20 x 4²

h₁ = 160 m

The second distance moved by the rocket is calculated as follows;

The velocity of the rocket before its thrust failed;

v = u + at

v = 0 +  20 x 4

v = 80 m/s

This becomes the initial velocity for the second stage

At maximum height, the final velocity = 0

[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]

The maximum height reached by the rocket = h₁ + h₂

                                                                          = 160 + 326.53

                                                                          = 486.53 m

if p=2i+4j+3k and q=I+5j-2k,find P×q.

Answers

Answer:

[tex]p\times q=-23i+7j+6k[/tex]

Explanation:

We are given that

p=2i+4j+3k

q=i+5j-2k

We have to find pxq

We know that

[tex]p\times q=\begin{vmatrix}  i&j  &k\\  2&4  & 3\\  1& 5 & -2\end{vmatrix}[/tex]

[tex]p\times q=i(-8-15)-j(-4-3)+k(10-4)[/tex]

[tex]p\times q=-23i+7j+6k[/tex]

Hence,[tex]p\times q=-23i+7j+6k[/tex]

A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water

Answers

Answer:

The speed of the swimmer in stil water is 0.5 m/s

Explanation:

Given;

total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s

width of the river, = 200 m

Please find the image attached for explanation.

A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .

Required:
How far from the second lens is the final image of an object infinitely far from the first lens?

Answers

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

 

The distance of the final image from the first lens will be is 6 cm.

What is mirror equation?

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

https://brainly.com/question/3229491

What is the relationship between organ systems and organs? organs are made from one type of organ system organ systems are made from one type of organ organs are made from different types of organ systems organ systems are made from different types of organs

Answers

Organs are made up of different types of organs.

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answers

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction

Answers

Answer:

The kinetic energy and potential energy lost to friction is 2,420 J.

Explanation:

Given;

total mass, m = 40 kg

initial velocity of the girl, Vi = 5 m/s

hight of the hill, h = 10 m

length of the hill, L = 100 m

initial kinetic energy of the girl at the top hill:

[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]

initial potential energy of the girl at the top hill:

[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]

Total energy at the top of the hill:

E = 500 J + 3920 J

E = 4,420 J

At the bottom of the hill:

final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s

hight of the hill = 0

final kinetic energy of the girl at the bottom of the hill:

[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]

final potential energy of the girl at the bottom of the hill:

[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]

Based on the principle of conservation of energy;  

the sum of the energy at the top hill = sum of the energy at the bottom hill

The energy at the bottom hill is less due to energy lost to friction.

[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]

Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.

g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Answers

Answer:

[tex]V=8.08m/s[/tex]

Explanation:

From the question we are told that:

Height[tex]h=5.00m[/tex]

Mass [tex]m=0.750kg[/tex]

Radius [tex]r=4.00cm=>0.04m[/tex]

Generally the equation for Total energy is mathematically given by

  [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]

Therefore

 [tex]V=\sqrt{\frac{4gh}{3}}[/tex]

 [tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]

 [tex]V=8.08m/s[/tex]

Help!!
A table is pushed across the floor for a distance of 32 m with a force of 320 N in 150 seconds. How much power was used?
A.70.2W
B.68.3W
C.56.7W
D.49.8W

Answers

Compute the work done on the table:

W = Fd = (320 N) (32 m) = 10,240 J

Divide this by the given time duration to get the power output:

P = W/∆t = (10,240 J) / (150 s) ≈ 63.3 W

which team won the champions league in 2020 2021​

Answers

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous

Answers

aluminum has the largest atomic radius

Answer:

francium

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

I need help with physics question.

Answers

(D)

Explanation:

Assuming that the charge q is moving perpendicular to the magnetic field B, the magnitude of the force experienced by the charge is

F = qvB = (2.9×10^-17 C)(4.0×10^5 m/s)(1.7T)

= 2.0×10^-11 N

If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?

Answers

Answer:

72  is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72  is the premimum of the insurance.

Define Potential Energy
Begin by defining potential energy in your own words within one concise eight word sentence

Answers

Answer:

potential energy is a type of energy an object has because of it's position

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time

Answers

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

you are given a set of facts regarding a lens : object heigh, and dostance to objects. Given this jnformation, how can you tell if you're dealing with a concave or convex lens

Answers

Answer:

concave curves inward like an hourglass and convex is an outward curve like a football

Explanation:

hope this helps

Problem

A charged particle is moving in the presence of uniform magnetic field. The mass of the particle

is m = 10−6 kg its charge is Q = 10−5 C and the magnetic field vector is B~ = (1T, 0, 0). At the

beginning the velocity vector of the particle is ~v0 = (12 m/s, 0, 5 m/s).

a.) How large will the x component of the velocity of the particle be in t = 2 s?

b.) Where will the particle be in t = 3.14 s?

c.) How large will the magnitude of the velocity be in t = 2.5 s?​

Answers

Answer:

Answer is a I checked the work

A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.

Answers

This question is incomplete, the complete question is;

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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

Answer:

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

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