2. If rain is falling vertically downward, and you are running for shelter, should you hold your umbrella
vertically, tilted forward, or tilted backward to keep the driest? Please explain.​

Answers

Answer 1

Answer:

Tilted forward to keep the driest.

Explanation:

The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.

The situation is the same as if you would stand still and the rain would come under an angle from the front.


Related Questions

small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? It is impossible to tell since the velocities are not given. The truck experiences the greater average force. It is impossible to tell since the masses are not given. The small car and the truck experience the same average force. The small car experiences the greater average force.

Answers

Answer:

The correct option is D: "The small car and the truck experience the same average force."

Explanation:

The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.

A ship can float on water as long as it weighs less than water.
O A. True
O B. False

Answers

Answer:

It's true

Explanation:

Because the ship is mafe up of aluminium, which is a light metal.

Answer:

False

Explanation:

Took The Quiz

A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.280 T/s.
A. What is the magnitude of the electric field induced in the ring?
B. In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?1. Counterclockwise2. Clockwise

Answers

Answer:

A. Ein = 8.05*10^-4 V/m

B. Clockwise sense

Explanation:

A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:

[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex]            (1)

Ein: induced electric field

ds: differential of a path of the ring

ФB: magnetic flux in the ring

The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):

[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex]   (2)

dB/dt = -0.280T/s     (it is decreasing)

A: area of the ring = π(r/2)^2= (π/4) r^2

r: radius of the ring = 4.60/2 = 2.30 cm

Then, you replace the values of all variables in the equation (2):

[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]

hence, the induced electric field is 8.05*10^-4 V/m

B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.

Which symbol is used to show vector quantities

Answers

Answer:  arrows

Explanation:

A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.

What is the vector quantity unit?

The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.

The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.

Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).

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Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and decrease their velocities at the same rate. If car B travels a distance D before stopping, how far does car A travel before stopping?
A) 4D
B) 2D
C) D
D) D/2
E) D/4

Answers

Answer:

A) 4D

Explanation:

The distance traveled by the cars before coming to rest can be determined by 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,

s = distance traveled

Vf = Final Speed = 0 m/s

Vi = Initial Speed

a = deceleration rate

First, we consider Car B and we assign a subscript 2 for it:

Vf₂ = 0 m/s  (As, car finally stops)

s₂ = D

a₂ = - a  (due to deceleration)

D = (0² - Vi₂²) /(-2a)

D = Vi₂²/2a    -------- equation (1)

Now, we consider Car A and we assign a subscript 1 for it:

Vf₁ = 0 m/s  (As, car finally stops)

s₁ = ?

a₁ = - a  (due to deceleration)

Vi₁ = 2 Vi₂  (Since, car A was initially traveling at twice speed of car B)

s₁ = (0² - Vi₁²) /(-2a)

s₁ = (2Vi₂)²/2a

s₁ = 4 (Vi₂²/2a)

using equation (1), we get:

s₁ = 4D

Therefore, the correct option is:

A) 4D

0.92 kg of R-134a fills a 0.14-m^3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a.

Answers

Answer:

The final volume of R-134a is 0.212m³

Explanation:

Using one of the general gas equation to find the final volume of the R-134a.

According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.

VαT

V = kT

k = V/T

V1/T1 = V2/T2 = k

Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K

V2 = ? at T = 100°C = 100+273 = 373K

On substituting this values for T2;

0.14/246.6 = V2/373

373*0.14 = 246.6V2

V2 = 373*0.14 /246.6

V2 = 0.212m³

The final volume of R-134a is 0.212m³

A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?

Answers

Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s

Explanation: Please see the attachment below

The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.

The given parameters:

Angular speed of the turn table = 1.85 rad/s²Time of motion, t = 4.0 sAngular displacement, θ = 30.0 rad

The angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;

[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]

where;

[tex]\omega_i[/tex] is the initial angular velocity

[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]

Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.

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A trap-jaw ant snaps its mandibles shut at very high speed, good for catching small prey. But these ants also slam their mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. If a 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms, at what speed will it leave the ground?

Answers

Answer:

FInal speed (v) = 0.509 m/s (Approx)

Explanation:

Given:

Mass of ant (m) = 12 mg

Force (f) = 47 N

Time taken (t) = 0.13 ms

Find:

FInal speed (v) = ?

Computation:

Initial velocity (u) = 0

Impulse = change in momentum

Force × TIme = change in momentum

47 × 0.13 = mv - mu

6.11 = 12 (V)

FInal speed (v) = 0.509 m/s (Approx)


Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
responsible for many processes and the habitable temperatures on the earth that
make our life possible.
a) Calculate the amount of energy arriving on the Earth in a single day
b) To how many litres of heating oil (energy density 37.3 x 10^6 J/litre is the equivalent?
C) The Earth reflects 30% of this energy : Determine the temperature on Earth's sufact
d) what other factors should be considered to get an even more precisa temperature postiache
Note: The Earth's radius is 6370km; the Sun's sadius is 696 ×10^3km, I AU is 1.495 × 10^8km)​

Answers

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = [tex]\sqrt[4]{P/Ae}[/tex]

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

For the parallel plates mentioned above, the DC power supply is set to 31.5 Volts and the plate on the right is at x = 14 cm. What is the magnitude of the electric field at a point on the x-axis where x = 7.0 cm? Answer with a number in the format ### in Newtons per Coulombs.

Answers

Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.

Answer:

E = 225 N/C

Explanation:

Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.

Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm

V(x) = 31.5 Volts

x = 14 cm = 0.14 m

The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:

E = V(x)/d

E(x = 0.14) = 31.5/0.14

E(x=0.14) = 225 N/C

E(x=0.14) = E(x=0.07) = 225 N/C

Organ pipe A, with both ends open, has a fundamental frequency of 475 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. Use 343 m/s for the speed of sound in air. How long are (a) pipe A and (b) pipe B?

Answers

Answer:

The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]

The length of organ pipe B is  [tex]L_b = 0.2708 \ m[/tex]

Explanation:

From the question we are told that

    The fundamental frequency is  [tex]f = 475 Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The fundamental frequency of the organ pipe A  is mathematically represented as

        [tex]f= \frac{v_s}{2 L}[/tex]

Where L is the length of  organ pipe

   Now  making L the subject

        [tex]L = \frac{v_s}{2f}[/tex]

substituting values

        [tex]L = \frac{343}{2 *475}[/tex]

        [tex]L = 0.3611 \ m[/tex]

The second harmonic frequency of the  organ pipe A is mathematically represented as

       [tex]f_2 = \frac{v_2}{L}[/tex]

The third harmonic frequency of the  organ pipe B is mathematically represented as      

      [tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]

So from the question

       [tex]f_2 = f_3[/tex]

So

    [tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]

Making  [tex]L_b[/tex] the subject

     [tex]L_b = \frac{3}{4} L[/tex]

substituting values

    [tex]L_b = \frac{3}{4} (0.3611)[/tex]

    [tex]L_b = 0.2708 \ m[/tex]

Explain why it is necessary to have a high voltage​

Answers

Answer:

SO THAT

EACH APPLIANCE CAN GET SUFFICIENT POTENTIAL DIFF. TO RUN

A car travels around an oval racetrack at constant speed. The car is accelerating:________.
A) at all points except B and D.
B) at all points except A, B, C, and D.
C) everywhere, including points A, B, C, and D.
D) nowhere, because it is traveling at constant speed.
2) A moving object on which no forces are acting will continue to move with constant:_________
A) Acceleration
B) speed
C) both of theseD) none of these

Answers

Answer:

1A,2D,3B

Explanation:

hope this helps

Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this

Answers

Answer:

Exothermic

Explanation:

Depending on the unit you are in, the answer may vary.

This is an exothermic reaction because it produces heat (the beaker gets warm).

g 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?

Answers

Answer:

Angle = 18.41°

Explanation:

Torque = F•r•sin θ

where;

F = force

r = distance from the rotation point

θ = the angle between the force and the radius vector.

We are given;

Torque = 15 N.m

F = 95 N

r = 0.5 m

Thus, plugging in the relevant values ;

15 = 95 × 0.5 × sin θ

sin θ = 15/(95 × 0.5)

sin θ = 0.3158

θ = sin^(-1)0.3158

θ = 18.41°

What is the answer for this question

Answers

ANSWER: My sister, who is a waitress at Billy’s Big Burger Shack, is sixteen years old.
The correct is c. If you need help with more questions you can dm me

Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?

Answers

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

 

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

A light wave will *Blank* if it enters a new medium perpendicular to the surface.

Answers

Answer:

A light wave will not stop if it enters a new medium perpendicular to the surface.

Explanation:

A light wave will not have any deviation if it enters a new medium perpendicular to the surface.

What is meant by refraction ?

Refraction is defined as an optical phenomenon by which the direction of a light wave gets changed when it travels from one medium to another. This is because of the change in speed.

Here,

The light wave is entering a new medium such that it enters perpendicular to the surface. Angle of incidence is the angle between the incident ray and the line perpendicular to the surface at the point of incidence. Since, here the light ray is incident normal to the surface that means the angle of incidence is 0.

According to Snell's law,

sin i = μ sin r

where i is the angle of incidence, r is the angle of refraction and μ is the constant called refractive index.

As i = 0, sin i = 0

So, μ sin r = 0

Since μ is a constant, we can say that sin r = 0 or the angle of refraction,

r = 0

This means that there is no refraction and hence the light wave won't get deviated when it enters the medium normally.

Hence,

A light wave will not have any deviation if it enters a new medium perpendicular to the surface.

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You are comparing a reaction that produces a chemical change and one that produces a physical change. What evidence could you use to determine which type of change is occurring?

Answers

Answer: A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.

Explanation: hope this helps have a good day

Answer:

If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.

A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.7 m from his television set. A reporter at the press conference is located 5.5 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.

Answers

Answer:

Explanation:

 Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.

5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343

d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.

.0160349 = d / 3 x 10⁸  + .0078717

d / 3 x 10⁸  = .0081632

d = 2448960 m

= 2448.96 km

= 2449 km .

What is The mass of an electron

Answers

9.10938356 × 10-31 kilograms

the part of the brain stem called the has been shown to related to arousal in lab animals. when this part is stimulated the animal is awake when it is severed cut the animal goes into coma

Answers

Answer:

Its called PSY

Explanation: I so do not know why they named it this way but, hope i helped.

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.37x10-2 kg/s. The density of the gasoline is 739 kg/m3, and the radius of the fuel line is 3.37x10-3 m. What is the speed at which gasoline moves through the fuel line

Answers

Answer:

Speed v = 2.04 m/s

the speed at which gasoline moves through the fuel line is 2.04 m/s

Explanation:

Given;

Mass transfer rate m = 5.37x10^-2 kg/s.

Density d = 739 kg/m3

radius of pipe r = 3.37x10^-3 m

We know that;

Density = mass/volume

Volume = mass/density

Volumetric flow rate V = mass transfer rate/density

V = m/d

V = 5.37x10^-2 kg/s ÷ 739 kg/m3

V = 0.00007266576454 m^3/s

V = 7.267 × 10^-5 m^3/s

V = cross sectional area × speed

V = Av

Area A = πr^2

V = πr^2 × v

v = V/πr^2

Substituting the given values;

v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))

v = 0.203678639672 × 10 m/s

v = 2.04 m/s

the speed at which gasoline moves through the fuel line is 2.04 m/s

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

Answers

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  [tex]Q =2.094 C[/tex]

Explanation:

From the question we are told that

    The diameter of the wire is  [tex]d = 0.205cm = 0.00205 \ m[/tex]

     The radius of  the wire is  [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]

     The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]

       The electric field change is mathematically defied as

         [tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]

     

Generally the charge is  mathematically represented as

       [tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]

Where A is the area which is mathematically represented as

       [tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]

 So

       [tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]

Therefore

      [tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]

substituting values

      [tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]

     [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]

From the question we are told that t =  5 sec

           [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]

          [tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]

         [tex]Q =2.094 C[/tex]

     

The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.

Given the following data:

Diameter of wire = 0.205 centimeters.Resistivity of aluminum = [tex]2.75\times 10^{-8}[/tex] Ohm-meters.[tex]E(t)=0.0004t^2-0.0001t+0.0004[/tex] Newton per coulomb.

Conversion:

Diameter of wire = 0.205 cm to m = 0.00205 meter.

Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]

To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:

First of all, we would find the area of the wire.

[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]

Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:

[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]

Where:

A is the area of a conductor.[tex]\rho[/tex] is the resistivity of a conductor.t is the time.E is the electric field.

Substituting the given parameters into the formula, we have;

[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]

[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]

When t = 5 seconds:

[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]

Q = 2.094 Coulomb.

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A particle with a charge of 5.1 μC is 3.02 cm from a particle with a charge of 2.51 μC . The potential energy of this two-particle system, relative to the potential energy at infinite separation, is

Answers

Answer:

U = 3.806 J

Explanation:

The potential energy between the two charges q1 and q2, is given by the following formula:

[tex]U=k\frac{q_1q_2}{r}[/tex]         (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = 5.1*10^-6 C

q2 = 2.51*10^-6 C

r: distance of separation between particles = 3.02cm = 3.02*10^-2 m

You replace the values of all parameters in the equation (1):

[tex]U=(8.98*10^9Nm^2/C^2)\frac{(5.1*10^{-6}C)(2.51*10^{-6}C)}{3.02*10^{-2}m}\\\\U=3.806J[/tex]

The potential energy of the two particle system is 3.806 J

Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.

Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?

Answers

Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

[tex]y_m=\frac{m\lambda D}{d}[/tex]    (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

You replace the values of the variables in the equation (1) for each wavelength:

For λ = 660 nm = 660*10^-9 m

[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]

For λ = 470 nm = 470*10^-9 m

[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]

In the Life Cycle of Stars diagram, what stage does letter J represent?
A.) white dwarf

B.) black dwarf

C.) black hole

D.) neutron star

Which letters in the Life Cycle of Stars diagram represent stars on the main sequence?

A.) F & I

B.) C & G

C.) A & E

D.) B & D

In the Life Cycle of Stars diagram, what stage does letter L represent?

A.) neutron star

B.) black hole

C.) white dwarf

D.) black dwarf

In the Life Cycle of Stars diagram, what stage does letter I represent?

A.) neutron star

B.) black dwarf

C.) black hole

D.) white dwarf

In the Life Cycle of Stars diagram, what does letter D represent?

A.) a high mass star

B.) a white dwarf

C.) a cool star

D.) a low mass star

In the Life Cycle of Stars diagram, what stage does letter C represent?

A.) nuclear fusion

B.) a supernova

C.) a planetary nebula

D.) protostar formation

Which letter in the Life Cycle of Stars diagram represents a star forming region of space?

A.) M

B.) H

C.) J

D.) G

Which letter in the Life Cycle of Stars diagram represents a planetary nebula?
Group of answer choices

A.) G

B.) H

C.) L

D.) M

Answers

ANSWER: num 1 is black hole

Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.

Answers

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)

W is described as: W = m.g

Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N

N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:

[tex]W_y[/tex] = m.g.cos(14)

Therefore, the formula will be:

[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy is 91.19 GeV, but its short lifetime shows up as an intrinsic width of 2.5 GeV. what is the lifetime of this particle?

Answers

Answer:

The lifetime of the particle is  [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Explanation:

From the question we are told that

    The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]

    The intrinsic width is  [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]

The lifetime is mathematically represented as

     [tex]\Delta t = \frac{h}{\Delta E}[/tex]

Where h is the Planck's constant with a value of  [tex]1.055*10^{-34} \ J\cdot s[/tex]  

substituting values

    [tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]

     [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

What is the goal of the Standing Waves lab? Group of answer choices To determine how frequency changes with mode number. To determine the velocity of a wave traveling on string. To determine wavelength of a wave on a string. To be the very best like no one ever was.

Answers

Answer:

To determine wavelength of a wave on a string.

Explanation:

The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.

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