The figure(Figure 1) shows the angular acceleration graph of a turntable that starts from rest. What is the turntable's angular velocity at t = 1.5 s? Express your answer to two significant figures and include the appropriate units.

Answers

Answer 1

The turntable's angular velocity at t = 1.5 s is 1.5 rad /s from the angular acceleration graph.

What is the relationship between angular velocity and acceleration?

When there is non-uniform circular motion, the rate of change in angular velocity (i.e., angular acceleration) is =t. = t. A t = v t denotes linear or tangential acceleration, which describes changes in velocity's magnitude but not its direction.

The turntable's angular velocity at t = 1.5 s

Angular acceleration( α )from the figure at t=1.5 s is 1 rad/s2.

      α => ω / t

      ω => α x t = 1 rad/s2 x 1.5 s

           => 1.5 rad /s

Are the directions of angular velocity and acceleration the same?

The angular acceleration points in the same direction as the angular velocity if the angular velocity is rising and growing. The angular acceleration points in the opposite direction of the angular velocity if the angular velocity is decreasing.

Learn more about angular velocity here:

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Related Questions

1. Clara stops for 10 minutes to catch up with a friend.

Answers

Answer:

Clara has speed of 80m/min

Explanation:

Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.

PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.



The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.


Which number label represents the cell membrane?


1

2

4

6

(this is middle school science)

Answers

Answer:

1. cell membrane

2. golgi body

3. mitochondrion

4. cytoplasm

5. nucleolus

6. nucleus

Explanation:

The correct answer to this question is Option A; 6.

Why?

In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.

~Thank you~

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.

Answers

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

[tex]V_{A}[/tex] = 1 m³

[tex]T_{A}[/tex] = 10°C = 283 K

[tex]P_{A}[/tex] = 350 kPa

[tex]m_{B}[/tex] = 3 kg

[tex]T_{B}[/tex] = 35°C = 308 K

[tex]P_{B}[/tex] = 150 kPa

Now, lets apply the ideal gas equation;

[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]

[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150

[tex]V_{B}[/tex] = 265.188 / 150  

[tex]V_{B}[/tex] = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, [tex]m_{A}[/tex] =  [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221

[tex]m_{A}[/tex]  = 4.309 kg

Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg

Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex]  = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

[tex]P_{f}[/tex] =  [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]

given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K

we substitute

[tex]P_{f}[/tex] =  ( 7.309 × 0.287 × 293)  / 2.77

[tex]P_{f}[/tex] =  614.6211119 / 2.77

[tex]P_{f}[/tex] =  221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move away from the object?

Answers

Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, [tex]\lambda=53\ cm=0.53\ m[/tex]

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

[tex]v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s[/tex]

So, the wave move with a speed of 7.95 m/s.

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.

Answers

Answer:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Explanation:

Given

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]

Point: [tex](f(t0), g(t0), h(t0))[/tex]

[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information

Required

Determine the parametric equations

[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]

Differentiate with respect to t

[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]

Let t = 1 (i.e [tex]t0 = 1[/tex])

[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]

[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]

[tex]r'(1) = i +\frac{1}{3}j + k[/tex]

To solve for x, y and z, we make use of:

[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]

This implies that:

[tex]r'(1)t = xi + yj + zk[/tex]

So, we have:

[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]

[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]

By comparison:

[tex]xi = it[/tex]

Divide by i

[tex]x = t[/tex]

[tex]yj = \frac{1}{3}jt[/tex]

Divide by j

[tex]y = \frac{1}{3}t[/tex]

[tex]zk = kt[/tex]

Divide by k

[tex]z = t[/tex]

Hence, the parametric equations are:

[tex]x = t[/tex]

[tex]y = \frac{1}{3}t[/tex]

[tex]z =t[/tex]

Can someone please help, ty!!
Will mark brainliest.

Answers

Answer:

4. unbalanced and Accelerating

5. balance and rest

Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].

One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.

a. Stretch
b. Vibrate
c. Contract
d. Expand

Answers

Answer:

The correct answer is option D.

Explanation:

It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.

I hope this helps.

A high-voltage direct-current generating station delivers 10 MW of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented by resistance RL and each of the two wires of the transmission line between the generating station and the city is represented by resistance RTL. The distance between the two locations is 2000 km and the transmission lines are made of 10 cm diameter copper wire. Determine (a) how much power is consumed by the transmission line and (b) 12 V I0 _

Answers

Answer:

The answer is below

Explanation:

The resistivity of copper is ρ = 1.72 * 10⁻⁸ Ωm, diameter d = 10 cm = 0.1 m

The resistance (R) of transmission line is given as:

Rtl = ρL / A; where ρ = resistivity of copper = 1.72 * 10⁻⁸ Ωm, L = length of transmission line = 2000 km = 2000000 m, A is the area of the wire = πd²/4 = π(0.1)²/4

[tex]R_{tl}=\frac{\rho L}{A}=\frac{1.72*10^{-8}*2000000}{\pi*0.1^2/4}=4.4 \ ohm[/tex]

Power = [tex]\frac{V_L^2}{R_L}[/tex]

Power = 10 MW = 10 * 10⁶ W

[tex]10*10^6=\frac{(250*10^3)^2}{R_L} \\\\R_L=\frac{(250*10^3)^2}{10*10^6} \\\\R_L=6250\ ohm[/tex]

[tex]I_L=\frac{V_L}{R_L} \\\\I_L=\frac{250*10^3}{6250} =40\ A[/tex]

a) Since there are two tranmission lines, the power consumed by the lines is:

[tex]P_{TL}=2*I_L^2*R_{TL}=2*40^2*4.4=14080\ W[/tex]

b) The energy generated by the source = 10 * 10⁶ W + 14080 W = 10014080 W

Fraction used = 10 * 10⁶ / 10014080 * 100% = 99.86%

what element is produced when a gold nucleus loses a proton?

Answers

The element is Platinum.
Hello, it’s Platinum.

An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0 cm width. The error in area (in cm) is​

Answers

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

Which example describes a nonrenewable resource?

Answers

Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.

At what tempreture will the of and oC
be the same​

Answers

Answer:

-40 degrees

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

Explanation:

Hope it is helpful.....

Answer:

To find the temperature when both are equal, we use an old algebra trick and just set �F = �C and solve one of the equations. So the temperature when both the Celsius and Fahrenheit scales are the same is -40 degrees.

a 2,400 kg car drives north towad a 60kg shopping cartthat has a velocity of zero the two objects collide giving the car a final velocity 4.33m/s north and the shopping cart 8.88m/s north what is the in itial velocity of the car

Answers

Answer:

4.552m/s

Explanation:

[tex]V=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1} } =\frac{2400*4.33+60*8.88}{2400}=4.552m/s[/tex]

The carts are moving on a level, frictionless track. After the collision all three carts stick together. Find the speed of the combined carts after the collision.

Answers

Answer:

0.13 m/s

Explanation:

Unfortunately, I don't have an explanation but I guessed the correct answer.


what is a vector quantity?

Answers

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

Vector Quantity

A physical Quantity, which has magnitude, direction and units But must follow the traingle law of vector addition

Example:- Force, velocity acceleration displacement etc.

What is the acceleration of a .3 kg mass when there is a net force of 25.9 N on it?

Answers

Answer:

86.33m/s^2

Explanation:

Acceleration = Force/Mass

= 25.9/0.3

= 86.33

The voltage v(t) = 141.4 cos (ωt) is applied to a load consisting of a 10Ω resistor in parallel with an inductive reactance XL=ωL = 3.77Ω. Calculate the instantaneous power absorbed by the resistor and by the inductor. Also calculate the real and reactive power absorbed by the load, and the power factor. Draw all the voltage, current and power waveforms, also the draw the circuit and phasor diagrams.

Answers

Answer:

A) P(t) = 2651.25 [ 1 - cos2wt ] W

B)  Real power = 999.79 watts

    Reactive power = 2652.86 VA

c) power factor = 0.3526

Explanation:

Given data:

V(t) = 141.4 cos (ωt)

R(t) = 10 Ω

Inductive reactance XL = ωL = 3.77 Ω

Ir(t) = V(t) / R(t) = 14.14

A) Calculate the instantaneous power absorbed by the resistor and by inductor

By resistor :

Pr(t) = V(t) * Ir(t) = 141.4 * 14.14 [tex]cos^{2} wt[/tex] = 1999.396 [tex]cos^{2} wt[/tex]

      hence Pr  = 999.698 (cos2ωt + 1) w

By Inductor :

Pl(t) = V(t) I'L(t) = 141.4 cosωt * 37.5 cos(ωt - 90)  

                        =  5302.5 [tex]sin^2 wt[/tex]

Hence Pl(t) = 5302.5 [tex]sin^2 wt[/tex]   w  =  2651.25 [ 1 - cos2wt ] W

B) calculate the real and reactive power

First we have to determine the power factor

Given that : V(t) = 141.4 cosωt  v ,   Ir(t) = 14.14 cosωt A

IL(t) = 37.5 cos (ωt - 90° )

The phasor representation of the above is :

V = [tex]\frac{141.4}{\sqrt{2} } <0^{0} v[/tex] = 141.4 ∠0° ,  Ir = 10 ∠ 0° , IL = 37. 5 ∠ -90°

Total load current = Ir + IL = 28.35 ∠ -69.35°

power factor = cos -69.35° = 0.3526

Next we will determine the Real and reactive power using the relation below

S = VI = 100 ∠ 0°  * 28.35 ∠ -69.35°

         = 2835 ∠ 69.35°

S = P + jQ = 999.79 + 2652.85 j

Real power  = 999.79 watts

Reactive power = 2652.85 VA

A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:

Answers

Complete question is;

A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:

A) 10 feet

B) 30 feet

Answer:

A) 0.728 ft/s

B) 1.8 ft/s

Explanation:

Let the the position of the worker in ft be denoted by s.

Since he begins to walk away at a constant rate of 3 ft/s, then;

ds/dt = 3 ft/s

Now, the rope will form a triangle, with width "s" and the height 40. Since distance from the connection point to the ceiling = 40 ft

Using pythagoras theorem, we can find the length of the rope on this side of the pulley.

Hence, the length of rope on this side of the pulley = √(s² + 40²)

Meanwhile, on the other side the length will be;

(80) - √(s² + 40²)

Also, height of the weight will be;

h = 40 - ((80) - √(s² + 80²))

h = √(s² + 80²) - 40

Differentiating this, we have;

dh/dt = (ds/dt) × (s/√(s² + 40²))

From earlier, we saw that ds/dt = 3 ft/s

Thus;

dh/dt = 3s/√(s² + 40²)

A) when he has walked 10 ft, it means that s = 10. Thus;

dh/dt = (3 × 10)/√(10² + 40²)

dh/dt = 0.728 ft/s

B) when he has walked 30 ft, it means that s = 30. Thus;

dh/dt = (30 × 3)/√(30² + 40²)

dh/dt = 1.8 ft/s

please help me I'm begging you Define and give examples of elements and compounds the structure of atoms (electrons, neutrons, and protons)

Answers

An element is a substance whose atoms all have the same number of protons: another way of saying this is that all of a particular element's atoms have the same atomic number. ... When atoms of the same element have different numbers of neutrons, they are called isotopes. Examples of Compounds:
Water - Formula: H2O = Hydrogen2 + Oxygen. ...
Hydrogen Peroxide - Formula: H2O2 = Hydrogen2 + Oxygen2 ...
Salt - Formula: NaCl = Sodium + Chlorine. ...
Baking Soda - Formula: NaHCO3 = Sodium + Hydrogen + Carbon + Oxygen3 ...
Octane - Formula: C8H18 = Carbon8 + Hydrogen18 Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged). The electron is a subatomic particle, symbol e⁻ or β⁻ , whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure.

In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position

Answers

Answer:

Explanation:

The relation between electric field and potential difference is as follows

E = - dV / dr

That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .

Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means

dV / dr = constant

E = constant

Electric field is constant .

So the option which is correct is

" points east and does not vary with position " .

A transformer has 150 turns in the primary coil and 350 turns in its secondary coil. If the primary coil has a voltage of 200 volts, how many volts will the secondary coil have?
242 volts
288
353
467

Answers

Answer:

467 volts

Explanation:

Vs/Vp = Ns/Np

Vs = Ns/Np × Vp

Vs = 350/150 × 200 = 7/3 × 200

Vs = 467 volts

[BWS.02]If the same experiment is repeated in different parts of the world by different scientists,

the results will be the same
the results will become invalid
the outcome of the experiment will be non testable
the outcome of the experiment will be non observable

Answers

Answer:

the results will be the same.it may be

If the same experiment is repeated in different parts of the world by different scientists, the results will be the same.

What is scientific experiment?

An experiment is a procedure that is carried out to support or refute a hypothesis, or to determine the efficacy or likelihood of something that has never been tried before. Experiments shed light on cause-and-effect relationships by demonstrating what happens when a specific factor is changed.

Controls are typically included in experiments to minimise the effects of variables other than the single independent variable. This improves the reliability of the results, often by comparing control measurements to the other measurements. Scientific controls are an essential component of the scientific method.

Hence, If the same experiment is repeated in different parts of the world by different scientists, the results will be the same.

Learn more about experiment here:

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The string will break if the tension in
it exceeds 0.180 N. What is the
smallest possible value of d (in cm)
before the string breaks?

Answers

Answer:

define d first?

you need to list more variables

Answer:

list more valuable unit

. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential

Answers

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.

Solution:

The relation between electric field and the electric potential is:

E = [tex]\frac{dV}{dr}[/tex]

So, making dV the subject, we have:

dV = E x dr

Integrating the above equation, we get.

V = [tex]\int\limits^_ {} \,[/tex]E x dr      Equation 1

Now, in 2-D

E is inversely proportional to the radius r.

E ∝ 1/r

So, we can write: replacing E ∝ 1/r in the equation 1

V ∝  [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr

Which implies that,

V ∝  log (r)

Hence, distance dependence expected for the electric potential =  ln (r)

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60-m-diameter blades at that location. Take the air density to be 1.25 kg/m3. Cengel, Yunus; Cengel, Yunus. Thermodynamics: An Engineering Approach (p. 98). McGraw-Hill Higher Education. Kindle Edition.

Answers

Answer:

1767Kw

Explanation:

Velocity of wind = 10 m/s

diameter of the blades= 60m

ρ= air density = 1.25 kg/m3

Acceleration due to gravity= 9.81 m/s^2

Mechanical energy of the wind can be calculated using the expression below

Energy= (e*m)

= ρ V A e............eqn(1)

Where A= area

ρ= air density

e= wind energy per unit mass of air

e= (v^2)/2..........eqn(2)

If we substitute the values into eqn (2) we have

e= [(10)^2]/2

=50J/Kg

But Area=A= (πd^2)/4

Area= ( π× 60^2)/4

Area=2827.8m^2

If we input substitute the values into eqn (1) we have

Energy= 1.25 ×10 × 50×2827.8

=1767145.7W

We can convert to kilo watt

=1767145.7W/ 1000

= 1767Kw

Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw

As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride

Answers

Answer:

Answer is explained in the explanation section below.

Explanation:

In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.

a) You pass through the highest point:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.

And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.

b) You pass through the lowest point of the ride:

When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.

And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.

In picture 1, heat is flowing from the ____ to the _____ In picture 2, heat is flowing from the _______ to the ____​

Answers

Answer: In picture 1, heat is flowing from the liquid to the air. In picture 2, heat is flowing from the air to the liquid

Explanation:

I don't know if I answered correctly, if not I can provide another answer

Fill in the question

Answers

4) 55m
5) 30 seconds
6) 1.83m/s

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these

Answers

Answer:

a) E = 0

b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

From which we have;

[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]

E = 0/A = 0

E = 0

b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]

[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]

[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]

By Gauss theorem, we have;

[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]

Therefore, we get;

[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]

The electrical field outside the spherical shell

[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]

[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]

Therefore, we have;

[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]

ALOT OF POINTS PLZ HURRYQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQWhat does Newton's third law say about why momentum is conserved in collisions?
A: Equal Forces act in equal times, so the change in momentum for both objects must be equal.
B: Unequal forces act for unequal times, so the change in momentum for both objects must be unequal.
C: Equal forces act for unequal times, so the change in momentum for both objects must be equal.
D: Unequal forces act for equal times, so the change in momentum for both objects must be equal.

Answers

Answer:

A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.

(Hope this helps! Btw, I am the first to answer.)

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