Fill in the blanks using the following words: Solid, Liquids, Gasses, More, Less, Gas, Fluids, Higher, Lower, Sun, Radiation, Conductors

● Radiation transfers heat best through _________ because there is _______ space between the particles.

● Conduction transfers heat best through _______ because there is space between the particles.

● Convection transfers heat best through _______ which includes _______ and ______.

● Heat always moves from _________ temperature to _________ temperature.

● Heat from ______ travels to earth by ___________.

● Solids that transfer heat well ate known as _____________.

​

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

Answer:

Explanation:

proton

Answer:

[tex]0.842\ \text{lb ft}[/tex]

[tex]0.1052\ \text{lb ft}[/tex]

Explanation:

d = Diameter of wheel = 6 in

r = Radius = 3 in = [tex]\dfrac{3}{12}=0.25\ \text{ft}[/tex]

t = Thickness = [tex]\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}[/tex]

w = Specific weight = [tex]425\ \text{lb/ft}^3[/tex]

[tex]t_2[/tex] = Time taken to slow down = 35 s

[tex]t_1[/tex] = Time taken to reach operating speed = 5 s

[tex]\omega[/tex] = Angular velocity = [tex]3450\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

Weight is given by

[tex]W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}[/tex]

Mass is given by

[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}[/tex]

Moment of inertia is given by

[tex]I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2[/tex]

Angular acceleration while slowing down is given by

[tex]\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2[/tex]

Frictional moment is given

[tex]\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}[/tex]

Frictional moment is [tex]0.1052\ \text{lb ft}[/tex]

Angular acceleration while speeding up is given by

[tex]\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2[/tex]

Motor torque is given by

[tex]\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}[/tex]

Motor torque is [tex]0.842\ \text{lb ft}[/tex].

Answer:

a = 52s²

Explanation:

How to find acceleration

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

Solve

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

t = -104 m/s / -2 m/s2

t = 52 s

Answer:

The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy

Answer:

Most likely (B)

Explanation:

B in the passage is the most representative out of all your choices and it has evidence from the passage  

Hope dis helps Jit!

Sorry i forgot to type C

B and C both measure mass while the others are calculations and are bias

The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:

A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.

When  a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.

When  a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.

Learn more about  gravitational field here:

https://brainly.com/question/26690770

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Answer:

nerves

Explanation:

I think, I maybe wrong.

Answer:

como  ías

Explanation:

Explanation:

Newton's second law of motion states that the external force is directly proportional to the rate of change of momentum. Mathematically, Newton's second law of motion is given by :

F = ma

Where

m is the mass and a is the acceleration

If there is a smaller mass, it would need a weaker force to accelerate it as the force is directly proportional to the mass. Hence, the correct option is (d).

Answer:

In absorption, the frequency of the incoming light wave is at or near the energy levels of the electrons in the matter. The electrons will absorb the energy of the light wave and change their energy state.

Explanation:

Answer:

E = 0.1472  J

Explanation:

Given that,

The number of turns in the solenoid, N = 2100

Area of the solenoid, A = 10 cm² = 0.001 m²

The length of the solenoid, l = 30 cm = 0.3 m

Current in the solenoid, I = 4 A

We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :

[tex]E=\dfrac{1}{2}LI^2[/tex]

Where

L is self inductance of the solenoid,

[tex]L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H[/tex]

So,

[tex]E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J[/tex]

So, 0.1472  J of energy is stored in the solenoid.

Answer:

You can use work to add kinetic energy to a system or to increase potential energy in the system.

Explanation:

Potential energy stored in any system can be released as kinetic energy. Kinetic energy can be transformed to do work or to increase potential energy.

hope this helped

Answer:

The de Broglie wavelength will remain the same because it does not depend on temperature.

Explanation:

de Broglie wavelength of a particle is independent of the temperature and hence the properties of emitted particle such as photoelectric effect, radioactive radiation etc. does not depend on the temperature.

Also, until unless the kinetic energy of a moving particle is not driven by the

thermal energy, the de Broglie wavelength  is independent of the temperature

Answer:

no work is done cause there is no movement of the wall

The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

A double replacement reaction have two ionic compounds that are exchanging anions or cations.

From the given options, we can choose the following based on their exchange of anions or cations.

Thus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.

Learn more about double replacement reaction here: https://brainly.com/question/14281077

#SPJ2

Answer:

475.7 m/s

Explanation:

Given,

Frequency ( f ) = 67 Hz

Wavelength ( λ ) = 7.1 m

To find : Speed ( v ) = ?

Formula : -

v = f λ

v

= 67 x 7.1

= 475.7 m/s

Therefore,

the speed of the wave is 475.7 m/s.

Answer: C. Stored energy

Answer: This is not easy lol

Explanation:

Answer:

Renewable energy minimizes carbon pollution and has a much lower impact on our environment. And it's having its moment in the sun. "Giving more New Yorkers access to renewable energy can allow them to reduce their own energy bills while reducing stress on the grid and demand for fossil fuel power.

Answer:

  k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

give us some initial conditions

1) friction force fr = 1N when v = 4m / s

2) an initial displacement of x = 0.08 m for t=0 s

Explanation:

In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law

Let's set a reference system with the y-axis in a vertical and positive direction upwards.

We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative

let's write Newton's second law

          F_e -F -fr - W = m a

where

          F_e = -kDy = - k y

          fr = - b v = -b dy / dt

          W = mg

we substitute for the specific case, that is, using the signs

          k y  -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]

In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero

         k y - m g - F = 0

from this equation you can find the spring constant, y= 9m and F=2 N

It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains

              k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

give us some initial conditions

1) friction force fr = 1N when v = 4m / s

2) an initial displacement of x = 0.08 m for t=0 s

therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.

             k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]

This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution

The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m

I believe the answer is a

Answer:

1992.008J

Explanation:

Answer: B

Explanation:

Answers A and C are examples of physical weathering while B is chemical weathering when water and lime mix it creates a reaction

Answer:

8 kV

Explanation:

Here is the complete question

Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?

Solution

Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V

So, Q = CV

= 500 × 10⁻⁶ F × 800 V

= 400000 × 10⁻⁶ C

= 0.4 C

Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV

V = Q/C

= 0.4 C/500 × 10⁻⁶ F

= 0.0008 × 10⁶ V

= 800 V

The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)

= 10 × 800 V

= 8000 V

= 8 kV

Answer:

[tex]1.62\times 10^{-8}\ \text{s}[/tex]

Explanation:

[tex]\epsilon_0[/tex] = Vacuum permittivity = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]

[tex]A[/tex] = Area = [tex]10\times 2\times 10^{-4}\ \text{m}^2[/tex]

[tex]d[/tex] = Distance between plates = 1 mm

[tex]V_c[/tex] = Changed voltage = 60 V

[tex]V[/tex] = Initial voltage = 100 V

[tex]R[/tex] = Resistance = [tex]1000\ \Omega[/tex]

Capacitance is given by

[tex]C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}[/tex]

We have the relation

[tex]V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}[/tex]

The time taken for the potential difference to reach the required level is [tex]1.62\times 10^{-8}\ \text{s}[/tex].