The electric field of a negative infinite line of charge: Group of answer choices Points perpendicularly away from the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and increases in strength at larger distances from the line charge Points perpendicularly away from the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge

Answer:

Newtons third law of motion: Balanced forces

Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs.   Once more, tug-of-war is a prime illustration.

The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.

The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.

Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.

Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.

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Answer:

C

Explanation:

The answer is C because if you look at the 1 hour mark it shows 10km

Answer:It will be 10km/hour

Explanation:

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 [tex]m^3[/tex]

The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]

Total load, [tex]m_{T}[/tex] = 290 kg

Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]

We known buoyant force,

[tex]$F_B = \rho_{air} V$[/tex]

For a 290 kg lift,  [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]

[tex]$m=\rho V$[/tex]

∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]

  [tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]

  [tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]

  [tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]

 ∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]

   [tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]

          [tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]

         = 545 K

          [tex]$=272^\circ C$[/tex]

Therefore, temperature of the air in the balloon is 272 degree Celsius.

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;

[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500

Therefore;

[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]

Therefore;

[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

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Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

Use formula:

Replace and solve:

The tension in the rope is 98 Newtons.

Greetings.

Answer:

81.1 m/s

Explanation:

The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration

So, T - f = ma

T -  μmg  = ma

a = T/m - μg

susbstμituting the values of the varμiables into the equation, we have

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 200 N/75 kg - 0.1 × 9.8 m/s²

a = 2.67 m/s² - 0.98 m/s²

a = 1.69 m/s²

Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s

So, v = u + at

v = 0 m/s + 1.69 m/s² × 48 s

v = 0 m/s + 81.12 m/s

v = 81.12 m/s

v ≅ 81.1 m/s

So, Jason's top speed is 81.1 m/s

The question is incomplete. The complete question is :

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.

Solution :

Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.

It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.

Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :

[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex]    and   [tex]$\phi_{B,f} = 0$[/tex]

Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,

[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]

  [tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]

  = 30.27 V

Therefore, the emf generated is 30.27 V.

Answer:

you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3

Answer:

hydrogen

Explanation:

hope it helps bye.

Answer:

[tex]\frac{1}{2}mR^2\omega^2[/tex]

Explanation:

The rotational kinetic energy of an object is given by [tex]KE_r=\frac{1}{2}I\omega^2[/tex], where [tex]I[/tex] is the object's moment of inertia/rotational inertia and [tex]\omega[/tex] is the object's angular speed.

What we're given:

Since no numerical value is given for any of these, it is implied the desired answer will be an equation in terms of the variables given.

Substituting [tex]I=MR^2[/tex]:

[tex]KE_r=\boxed{\frac{1}{2}mR^2\omega^2}[/tex]

Answer:

[tex]Weight\ loss=1.6321N[/tex]

Explanation:

From the question we are told that:

Weight [tex]W=85.9kg[/tex]

Altitude [tex]h= 6.33 km[/tex]

Let

Radius of Earth [tex]r=6380km[/tex]

Gravity [tex]g=9.8m/s^2[/tex]

Generally the equation for Gravity at altitude is mathematically given by

 [tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]

 [tex]g_s=9.781m/s^2[/tex]

Therefore

Weight at sea level

 [tex]W_s=9.8*85.9[/tex]

 [tex]W_s=841.82N[/tex]

Weight at 6.33 altitude

 [tex]W_a=9.781*85.9[/tex]

 [tex]W_a=840.2N[/tex]

Therefore

 [tex]Weight loss=W_s-W_b[/tex]

 [tex]Weight loss=841.82-840.2[/tex]

 [tex]Weight loss=1.6321N[/tex]

Answer:

Explanation:

the force of the rocket engine pushing it up,  the force of gravity pulling it down,    maybe some force of air resistance as the rocket goes fast,   hmmm    Free Body Diagrams  (FBD)  should have any and all forces on the model,  unless they are negligible . or so slight they really make little difference in the total  outcome.  

Answer:

Explanation:

The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.

Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

Answer:

Explanation:

His Kinetic energy = 1/2 m v^2

v = 5 m/s

m = 75 kg

Ke = 1/2 75 * 5^2

Ke = 937.5 Joules

This will be converted to PE when he reaches the maximum height he reaches. In other words KE = PE

PE = m * g * h

m = 75

g = 9.81

h = ?

PE = 937.5

937.5 = 75 * 9.81 * h

937.5 = 735.75 * h

937.5/735.75 = h

h= 1.27 meters

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

∴

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s

The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.

The force is the action of push or pull which makes an object to move or stop.

Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by

F = m(g-a)

F = 13.9 x (9.81 - (-0.898))

F = 148.84 N

Thus, the magnitude of the force is  148.84 N.

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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. It must stay the same, but kinetic energy (KE) can be transformed to PE and PE can be transformed to KE within the system.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Potential energy (PE): it is an energy possessed by an object or body due to its position above the earth.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where,

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

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

Where;

Furthermore, the total energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Total energy = P.E + K.E

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

In this scenario, a system has both potential energy (PE) and kinetic energy (KE).

According to the law of conservation of energy, we can infer or deduce that the total energy of the system must stay the same because it cannot be destroyed, but kinetic energy (KE) can be transformed to potential energy (PE) and potential energy (PE) can be transformed to kinetic energy (KE) within the system.

Answer:

me...:(

Explanation:

Answer:

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

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]

Explanation:

4. Alin sa mga sumusunod na awitin ang may tempong presto?

a. “Chua-ay”

c. “Akong Manok”

b. “Sitsiritsit

d. "Ili-ili Tulog Anay”

5. Pakinggan ang awiting “Sa Ugoy ng Duyan” Ano ang tempo nito?

a. mabagal

c. mabilis na mabilis

b. mabilis at mabagal d. katamtamang bilis

6. Alin sa sumusunod na elemento ng musika ang nakikilala sa pamamagitan ng pakikinig o pag-awit na may nipis at kapal na tunog?

a. descant

b. ostinato

c. tempo

d. texture

7. Alin sa 2-part vocal ang nasa ibabang bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

8. Alin sa 2-part vocal ang nasa itaas na bahagi ng musical score?

a. alto

b. forte

c. tempo

d. soprano

9. Ano ang tawag sa paulit-ulit na rhythmic pattern na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

10. Ano ang tawag sa paulit-ulit na rhythmic pattern at may kasamang melody na ginagamit sa pansaliw ng awitin?

a. descant

b. ostinato

c. melodic ostinato

d. rhythmic ostinato

Answer:

A. the change in smell

Explanation:

sry if its wrong it's what I know

Answer:

both the same

Explanation:

When a ball is thrown vertically upwards, it experiences that same acceleration due to gravity as an object thrown directly downwards.

This means that if we ignore the effects of air resistance, and the two balls have the same initial speed, they are expected both expected to hit the ground at the same speed as a result of the principle of conservation of energy.

Answer:

Both balls will hit the ground at the same speed because air resistance is ignored.

Explanation:

Answer:

a)I=V/R

39.5 amp

Explanation:

because the voltage in serious with 1ohm resistor

Answer:

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Explanation:

Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v

As we know,  

m*v = Integration of F.dt with limits 0 to T

Ft = mv

F = mv/t

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Answer: 10 J

Explanation:

Trust ;)

Answer:

The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs

The pair of forces described by Newton third law must be in opposite direction.

Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.

The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.

When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.

A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.

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