The Heat Force
이
18
1 point
-
If two objects are the same temperature and are physically touching which of the following would be true?
The objects would be in thermodynamic equilibrium and would transfer energy through conduction.
ОООО
1
The objects would not be in thermodynamic equilibrium and heat would transfer through conduction.
The objects would not be in thermodynamic equilibrium and as a result there would be no heat transfer
The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.
2

Answer:

the number of electrons should equal to the the number of protons in a neutral atom

if there is a inequality between the numbers it means the atom has a + or - charge

Answer:

the speed in -y

Explanation:

For this exercise we must use the right hand rule. The motion of a positive charge is given by.

Thumb points in the direction of speed

fingers extended in the direction of the magnetic field, + z axis

the palm in the direction of the force, as the charge is negative in the opposite direction of the force, axis + x

therefore the thumb is in the direction - y

the speed in -y

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

Given the following data;

Radius, r = 2.6 km

Time = 360 seconds

Conversion:

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]

Where;

Substituting into the formula, we have;

[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]

[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]

Where;

Substituting into the formula, we have;

[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]

[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]

Centripetal acceleration = 0.79 m/s²

Answer:

B) Inertia is the resistance of any physical object

Answer: [tex]56.4\ m/s^2, 201.42\ m/s^2[/tex]

Explanation:

Given

Rocket attain a velocity of [tex]v=282\ m/s[/tex] in a time period of [tex]t=5\ s[/tex]

It was brought jarringly back to rest in only [tex]t'=1.4\ s[/tex]

Acceleration is the change in velocity of the object over a period of time

(a) Acceleration in his direction of motion

[tex]\Rightarrow a=\dfrac{v-0}{t}\\\\\Rightarrow a=\dfrac{282}{5}\\\\\Rightarrow a=56.4\ m/s^2[/tex]

(b) acceleration opposite to his direction of motion i.e. deceleration is

[tex]\Rightarrow a_d=\dfrac{0-v}{t'}\\\\\Rightarrow a_d=\dfrac{-282}{1.4}\\\\\Rightarrow a_d=-201.42 \ m/s^2\\\Rightarrow a_d=201.42\ \text{decelration}[/tex]

Answer: [tex]4.65\ m/s[/tex]

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

[tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]

So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]

Answer:

Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. ... During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.

Explanation:

thank me later

Answer:

[tex](b)\ t_1 - t_0[/tex]

[tex](d)\ t_2 - t_1[/tex]

[tex](e)\ \frac{t_2 - t_0}{2}[/tex]

Explanation:

Given

See attachment for complete question

Required

How long to reach the ground from the maximum height

First, calculate the time of flight (T)

[tex]T =t_2 - t_0[/tex]

The time taken (t) from maximum height to the ground is:

[tex]t = \frac{1}{2}T[/tex]

So, we have:

[tex]t = \frac{t_2 - t_0}{2}[/tex]

Another representation is:

At ymax, the time is: t1

On the ground, the time is t2

The difference between these times is the time taken.

So;

[tex]t = t_2 - t_1[/tex]

Since air resistance is to be ignored, then

[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height

Answer: [tex]14.64\ N[/tex]

Explanation:

Given

Inclination of ramp is [tex]\theta=15^{\circ}[/tex]

Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal

Weight of cart [tex]W=40\ lb[/tex]

from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp

Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position

[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]

speed = 40 m/s

Explanation:

Since the object is dropped, V0y = 0.

Vy = V0y - gt

= -(10 m/s^2)(4 s)

= -40 m/s

This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.

The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,

By using the first equation of motion,

v = u + at

initial velocity(u) = 0 m/s

acceleration(a) = 10 m/s²

v = 0 + 10×4

v = 40 meters/seconds

Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.

Learn more about equations of motion from here,

brainly.com/question/5955789

#SPJ2

Answer:

me...:(

Explanation:

Answer:

hello I'm online here thanks for the points (◔‿◔)

Force is mass into acceleration

and pressure is force applied per unit area.

Answer:

kinetic energy of the train = 2,910.6 x 10⁷ joule

Explanation:

Given:

Mass of train = 3.3 x 10⁷ kg

Speed of train = 42 m/s

Find:

kinetic energy of the train

Computation:

kinetic energy = (1/2)(m)(v²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)

kinetic energy of the train = (3.3 x 10⁷)(882)

kinetic energy of the train = 2,910.6 x 10⁷ joule

Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].

Explanation:

Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]

Speed = 42 m/s

Kinetic energy is the energy acquired by an object due to its motion.

Formula to calculate kinetic energy is as follows.

[tex]K.E = \frac{1}{2}mv^{2}[/tex]

where,

m = mass of object

v = speed of object

Substitute the values into above formula as follows.

[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]

Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].

Answer:

Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge

Explanation:

The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.

If λ is negative, E is negative so it points in the negative direction towards the line of charge.

Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.

Also as r increases, E decreases since E ∝ 1/r

So, the electric field decreases at larger distances from the line of charge.

So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.

Answer:

speed = 0.9 mm/s

Explanation:

time, t = 40 s

initial angular speed, wo = 0 rad/s

final frequency, f = 1/1.03 rps = 0.97 rps

final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s

time, t = 40 s

distance, r = 5.9 mm

The angular acceleration is given y the first equation of motion.

[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]

The linear velocity is

[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]

speed, v = 0.9 mm/s

Answer:

(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J

(b) the efficiency of the athlete is 10.94 %

Explanation:

Given;

work done by the athlete (system), W = 6.53 x 10⁴ J

the heat given off by the athlete (system), Q = 5.97 x 10⁵ J

The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.

                            Q← ⊕ →W

The work, W, points away from the system since the system does the work

The heat, Q, points away from the system since heat is given off

Apply first law of thermodynamic;

ΔU = Q + W

where;

q is the heat flowing into or out of the system

(+q     if the heat is flowing into the system

(-q      if the heat is leaving the system

w is the work done by or on the system

(+w     if the work is done on the system by the surrounding

(-w     if the work is done by the system to the surrounding

Thus, from the above explanation, the change in internal energy of the system is calculated as;

ΔU = -Q - W

ΔU = - 5.97 x 10⁵ J  -  6.53 x 10⁴ J

ΔU = -6.623 x 10⁵ J

The magnitude of the change in internal energy = 6.623 x 10⁵ J

(b) the efficiency of the athlete;

[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]

Answer:

4.408 [tex]\mathsf{M_{sun}}[/tex]

Explanation:

According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:

[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]

[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]

combined mass (m₁+m₂)  = 4.408 [tex]\mathsf{M_{sun}}[/tex]

Answer: The correct statements are:

Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

The key point about solid are:

Answer:

B. carry a single circuit and are placed in individual cases.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

Similarly, an integrated circuit (IC) also referred to as microchip can be defined as a semiconductor-based electronic component that comprises of many other tiny electronic components such as capacitors, resistors, transistors, and inductors.

Integrated circuits (ICs) are often used in virtually all modern electronic devices to carry out specific tasks or functions such as amplification, timer, oscillation, computer memory, microprocessor, etc.

A wafer can be defined as a thin slice of crystalline semiconductor such as silicon and germanium used typically for the construction of an integrated circuit.

In an integrated circuit, each wafer is cut into sections, which generally comprises of a single circuit that are placed in individual cases.

Additionally, a semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity.

Answer: B got it right on the test just now

Explanation:

Answer: 10 J

Explanation:

Trust ;)

Answer:

We should not try to park the car because its rest length is greater than the space available.

The car seems to be approximately equal to the friend (L = 3.16 m). Due to this reason he is sure to park.

Explanation:

We should not try to park the car because its rest length is greater than the space available.

The friend is sure about parking because the car appears short in length to him. For this, we will solve Einstein's length contraction formula from theory of relativity:

[tex]L = L_o\sqrt{1-\frac{v^2}{c^2}}[/tex]

where,

L = Relative length observed by friend = ?

L₀ = rest length = 6 m

v = relative speed = 85% of speed of light = 0.85c

Therefore,

[tex]L = (6\ m)\sqrt{1-\frac{(0.85c)^2}{c^2}}[/tex]

L = 3.16 m

Hence, the car seems to be approximately equal to the friend. Due to this reason he is sure to park.

Answer:

2.00x 10 14th Hz

Explanation:

Answer:

2.99 x 10^14 Hz

Explanation:

E photon= hf (you have to solve for f)

f= E photon/h

f= 1.98 x 10^-19 J / 6.63 x 10^-34 J x s

f=2.99 x 10^14 Hz

You decide to play fetch with your dog, who is sitting nextto you, so you throw a ball down a narrow hallway. The ballcomes to a stop 3.9 m down the hallway. The dog, startingfrom rest, runs after the ball with a constant acceleration of0.70m/s2until she reaches the ball. She grabs the ball whilestill running down the hallway uniformly accelerating(slowingdown) for 4.7 more seconds until she comes to a stop. What isthe total distance the dog travels to grab the ball and come toa final stop, starting from rest

Answer:

Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.

Explanation:

Complete question:

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.

Answer:

The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

Explanation:

Given;

power of the flux produced, P = 1 W = 1 J/s

energy per photon, E = 78 keV

Convert the energy per photon to J

E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon

let the number of photons = n

n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s

[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]

Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

Answer:

Y = 5.03 x 10⁻³ m = 5.03 mm

Explanation:

Using Young's Double-slit formula:

[tex]Y = \frac{\lambda L}{d}[/tex]

where,

Y = Fringe Spacing = Width of bright fringe = ?

λ =  wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = Screen distance = 3.1 m

d = slit width = 0.39 mm = 3.9 x 10⁻⁴ m

Therefore,

[tex]Y = \frac{(6.33\ x\ 10^{-7}\ m)(3.1\ m)}{3.9\ x\ 10^{-4}\ m}[/tex]

Y = 5.03 x 10⁻³ m = 5.03 mm