What is this sport ⚽⚾

Answer:

D) all of the above.

Explanation:

First polarises it, cahrging and discharging occurs at once.

A negatively charged balloon touching a wooden wall then from the given options option D is correct which is all of the above.

Charged matter experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. Positive or negative charges can exist in an electric field (commonly carried by protons and electrons, respectively).

Contrary charges attract one another, while like charges repel one another. A neutral object is one that carries no net charge. Classical electrodynamics, the name given to an early understanding of how charged particles interact, is still accurate for issues that do not call for taking into account quantum phenomena.

In the first step it polarizes molecules in the wall, then charging and discharging in the wall will take place at once.

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Answer:

122.5 ohm

Explanation:

Given :

P=40 w

V= 70 V

R=?

Resistance can be calculated as :

[tex]P=\frac{V^{2} }{R} \\40=\frac{(70)^{2} }{R}\\40=\frac{4900}{R} \\R=\frac{4900}{40} \\R=122.5 ohm[/tex]

Therefore, resistance of the bulb will be 122.5 ohm

Answer:

if i am not mistaken the volume is 7, because it only took that much space

Answer:

Explanation:

Since the x and y components are given

The vectors Magnitude = √32²+(-59)²

=67.12m

Solution :

The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]

So the distance appeared to Bob due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]

    [tex]$=7.1 \ Ly$[/tex]

Therefore, the time required to finish the race by Bob is

[tex]$t_b = \frac{L_b}{V_b}$[/tex]

  [tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]

So the distance appeared to Charlie due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]

    [tex]$=5 \ Ly$[/tex]

The time required to finish the race by Charlie is

[tex]$t_b = \frac{L_c}{V_c}$[/tex]

  [tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]

  = 5.77 year

Answer:

the correct answer is B

Explanation:

Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved

instantly starts. Before the crash

         p₀ = M v +0

final instant. After the crash

        m_f = (M + M) v_f

the moment is preserved

        p₀ = p_f

        M v = 2 M v_f

        v_f = v / 2

let's look for kinetic energy

before the crash

       K₀ = ½ M v²

after the crash

       K_f = ½ 2M (v_f)²

       K_f = ½ 2M (v/2)²

       K_f = (½ M v²) ½

       K_f = K₀ / 2

therefore the correct answer is B

Answer:

Boyle's law.

Explanation:

Robert Boyle was an Irish chemist and is famously referred to as the first modern chemist. He was born on the 25th of January, 1627 in Lismore, Ireland and died on the 31st, December 1691, London, United Kingdom.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

[tex] PV = K[/tex]

[tex] P_{1}V_{1} = P_{2}V_{2} [/tex]

Where;

P1 is the original pressure.

P2 is the final pressure.

V1 is the original volume.

V2 is the final volume.

Hence, when you release some of the paint from a spray paint can by applying an amount of pressure and the can remains at the same temperature, the gas law which this represent is Boyle's law.

1) (40 N) cos(30°) ≈ 34.6 N

2) (40 N) sin(30°) = 20 N

3) (65 kg) g = (65 kg) (9.80 m/s²) = 585 N

4) The net force on the sled acting in the vertical direction is made up of

• the sled's weight, 585 N, pointing downward

• the vertical component of the applied force, 20 N, pointing upward

• the normal force, with magnitude n, also pointing upward

The sled does not move up or down, so by Newton's second law,

∑ F = n + 20 N - 585 N = 0   ==>   n = 565 N

5) The net force in the horizontal direction consists of

• the horizontal component of the applied force, 34.6 N, acting in the direction the sled's movement (call this the positive direction)

• kinetic friction, with magnitude f, pointing in the opposite and negative direction

By Newton's second law,

∑ F = 34.6 N - f = 0   ==>   f ≈ 34.6 N

Now if µ is the coefficient of kinetic friction, then

f = µn   ==>   µ = f/n = (34.6 N) / (565 N) ≈ 0.0613

The component of the force is the effective part of that force in that direction.

The component of the force is the effective part of that force in that direction.

1) The horizontal component of a force = 40 N cos 30 degrees = 34.6 N

2) The vertical component of the force = 40 N sin 30 degrees = 20 N

3) The magnitude of the gravitational force = mg cos 30 degrees  = 65 Kg * 9.8 m/s^2 * cos 30 degrees = 551.7 N

4) The normal force = 551.7 N

5) The coefficient of friction = F/R =  40 N /551.7 N = 0.07

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1517.4 m

Step-by-step explanation:

Since the projectile broke up at the peak of its flight, it already traveled half its initial range so we can find its initial launch velocity [tex]v_0[/tex] from the equation

[tex]\frac{1}{2}R= \dfrac{1}{2} \left(\dfrac{v_0^2}{g}\sin 2\theta_0 \right)[/tex]

where [tex]\theta_0 = 45°[/tex] and [tex]\frac{1}{2}R = 50\:\text{m}[/tex] so we will get [tex]v_0=31.3\:\text{m/s}[/tex]. Next, we can use the equation

[tex]v_y = v_0y - gt = v_0 \sin 45 - gt[/tex]

and since [tex]v_y=0[/tex] at its peak, we get t = 22.1 s. Let's set this aside for a moment and we'll use it later.

At the top of its peak, we can use the conservation law of linear momentum. Let M be the mass if of the original projectile, [tex]m_1[/tex] be the mass of the larger fragment (2 kg) and [tex]m_2[/tex] be the mass of the smaller fragment (1 kg). We can write the conservation law as

[tex]Mv_0x = m_1V_1 + m_2V_2[/tex]

where [tex]V_1\:\text{and}\:V_2[/tex] are the velocities of the fragments immediately after the break up. But we also know that [tex]V_1=0[/tex] so the velocity of [tex]m_2[/tex] can be calculated from the conservation law as

[tex]Mv_0 \cos 45° = m_2V_2[/tex]

or

[tex]V_2 = \dfrac{M}{m_2}v_0 \cos 45° = 66.4\:\text{m/s}[/tex]

Now we can calculate the horizontal distance the smaller fragment traveled after the break up. Recall that the amount of time for it to go up is also the amount of time to get down so the horizontal distance x is

[tex]x = V_2 t = (66.4\:\text{m/s})(22.1\:\text{s})= 1467.4\:\text{m}[/tex]

Therefore, the total distance traveled from the launch point is

[tex]D = 50\:\text{m} + 1467.4\:\text{m}=1517.4\:\text{m}[/tex]

Answer:

jsgssbvwsvs

Explanation:

ifmd understand

Answer:

below

Explanation:

Part A) This engine works  per cycle is 254.9 J.

Part B) The thermal efficiency is  23.42%

The thermal efficiency of any heat engine is represented in percentage of heat energy converted into work.

For isothermal expansion, work done is

W₁ =nRT₁ x ln(V₂/V₁)

W₁ = 0.1 x 8.314 x 800 x ln(5000/3000)

W₁ = 339.8 J =Q₁

For isochoric cooling ,

W₂ =0

Q₂ =nCvdT = 0.1 x 3R/2 x (T₂-T₁)

Q₂ = -748.3 J

For isothermal compression,

W₃ =nRT₂ ln (V₄/V₃)

W₃  = 0.1 x 8.314 x 200 x ln(3000/5000)

W₃ = -84.9J

For isochoric heating

W₄ =0

Q₄ =nCvdT = 0.1 x 3R/2 x (800-200)

Q₄  = -748.3 J

Total work done in all the process W = W₁ +W₂ +W₃ +W₄

W =254.9 J

Thus, the work done is 254.9 J

Thermal efficiency = Work done/Heat taken

η = W/ Q₁ +Q₄

η = [254.9 / 339.8 +748.3 ] x 100 %

η = 0.2342 x 100 %

η = 23.42%

Thus, the thermal efficiency is 23.42%

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Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

Answer:

because density of iron is more than that of water but less then that of Mercury

hope it's helpful

Answer:

hey. i dont know what you tryna say but if u replying to someone else, you should use the comments section. in that way you won't lose points.

Answer:

17280 J and 1080 J

Explanation:

Given :

R= 40 ohm

I=1.2A

t= 5 min=60×5=300 sec

Now,

Total energy can be calculated as:

[tex]E=I^{2} Rt\\E=(1.2)^{2} *40*300\\E=17280 J[/tex]

Now,

V=12V

R=40 Ohm

[tex]E=\frac{V^{2} }{R} *t\\E=\frac{(12)^{2} }{40} *300\\E=1080 J[/tex]

Total energy  is 17280 J and 1080 J

Answer:pp

Explanation:

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Answer: Why am I growing is a chemical change.

Explanation:

Changes that can be reversed and does not affect the composition of a substance are called physical changes.

For example, change in state of a substance like ice converting into water is a physical change.

Changes that cannot be reversed and affect the chemical composition of a substance are called chemical changes.

For example, a child growing is an irreversible change and hence, it is a chemical change.

Thus, we can conclude that why am I growing is a chemical change.

Answer:

A wave model of light is useful for explaining brightness,color, and the frequency-dependent bending of light at a surface between media. However, because light can travel through space, it cannot be a matter wave, like sound or water waves.

Explanatio

omega=2pi/T

Answer:

0

0000

Explanation:

Answer:

British East India Company

The battle was fought at Buxar, a "small fortified town" within the territory of Bihar, located on the banks of the Ganga river about 130 kilometres (81 mi) west of Patna; it was a decisive victory for the British East India Company.

Answer:

Total probability = 3/8

Explanation:

Below is the calculation:

Number of molecules in the container = 3

The probability of one molecule in the left half, P = 3 / 2 = 1.5 or 1/2

The probability of second molecule in the left half, P1 = (3/4)

Total probability = P x P1

Total probability = (1/2) x (3/4)

Total probability = 3/8

Answer: Members of the Flat Earth Society claim to believe the Earth is flat. Walking around on the planet's surface, it looks and feels flat, so they deem all evidence to the contrary, such as satellite photos of Earth as a sphere, to be fabrications of a "round Earth conspiracy" orchestrated by NASA and other government agencies.

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two electrons are created from a quanta of kight

Answer:

B = 4.51×10⁻⁵ T

Explanation:

Given that,

The average magnetic field strength of the Earth is 0.0000451 T.

We need to write the value in the form of scientific notation. Any number in scientific notation is written as follows :

N=a×bⁿ

Where

n is any integer and a is a real no

So,

0.0000451 = 4.51×10⁻⁵ T

So, the required answer is equal to 4.51×10⁻⁵ T.

Answer:

below

Explanation:

Answer:

To produce a single sugar molecule a plant must use a molecule of water (H2O) and carbon dioxide (CO2) and sunlight. To get 4 molecules, you simple use 4 molecules of water and 4 molecules of carbon dioxide

Answer:

Sentences with many clauses and phrases are difficult to understand because the clauses and phrases typically _____.

modify other clauses and phrases in the sentence

refer to other sentences in the passage

make it hard to determine where the sentence ends

change the intended meaning of the sentence

Explanation:

Explanation:

1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:

[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]

[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]

Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so

[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]

Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so

[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]

Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is

[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]

2. The total current in the circuit is

[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]

3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is

[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]

4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us

[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]

5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is

[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]

This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is

[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]