Plzzz answer this question correctly

Answer:

(A) By reducing friction

Answer:

A

Explanation:

The solution to the questions are given as

Given, $B(t)=(1.4 T) e^{-0.057 t}$

[tex]$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}[/tex]

[tex]\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$[/tex]

[tex]\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}[/tex]

[tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]

(b) [tex]Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$[/tex]

[tex]\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}[/tex]

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule.  The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.

To compute the Molecular Mass of [tex]$O_{3}[/tex]

Atomic mass of oxygen(O) = 16

As [tex]$O_{3}[/tex] contains 3 atoms,

The molecular mass of [tex]$O_{3}[/tex]

= (16 x 3) = 48g/mol

Hence the mass of 9.11 moles O3

= 9.11 mol x 48g/mol

= 437.28g

= 0.43728kg

Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg

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Answer:

Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.

Explanation:

From the calculation, the atomic weight of the boron atom is 10.812.

We can find the atomic weight of boron from;

(19.78/100 * 10.0129) + ( 80.22/100 * 11.00931)

= 1.981 + 8.831

= 10.812

The number of protons, neutrons, and electrons in the following isotopes are;

N - 7 electrons, 7 protons and 7 neutrons

Cl - 17 electrons, 17 protons and 18 neutrons

Ca - 20 protons, 20 electrons and 28 neutrons

Cu - 29 protons, 29 electrons and 3 neutrons

U - 92 protons, 92 electrons and 138 neutrons

The number of outer shell electrons in the atoms are;

Na - 1

P - 5

Br - 7

I - 7

Te - 6

The elements are classified as follows;

Radon - nonmetal

Francium - metal

Arsenic - semiconductor

phosphorus - nonmetal

Hafnium - metal

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Answer:

FRICTION

Explanation:

Friction is a force, the resistance of motion when one object rubs against another.

Frictional force

Explanation:

Its the opposing force against horizontal motion

The confidence interval is given as lower interval is -0.77 while The upper interval is 1.67

Multivitamin treatment

n1 is 10

σ1 = 1.8

x21 = 5

vitamin C treatment

n2 = 10

σ2 = 1.5

x2 = 4.55

zα/2 = 1.645

The formula for the confidence interval is given as

(x1 - x2) ± zα/2[tex]\sqrt{\frac{sd1^2}{n1}+\frac{sd2^2}{n2} }[/tex]

When we input the values we have above we would have

CI = (-0.7688550 , 1.668855 )

The the lower interval is -0.77

The upper interval is 1.67

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A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 10 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 10 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained:

Treated with multivitamin

Days to recover from flu

2.4, 6.4, 9.1, 4.1, 4.6, 6.4, 6.4, 3.2, 6.9, 0.5

Treated with Vitamin C

5.2, 3, 3.6, 5.5, 7.5, 6.7, 1.3, 1.9, 5.3, 5.5

Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the

population standard deviation of recovery time from the flu is 1.5 days when treated with vitamin C tablets. Suppose also that both populations are

approximately normally distributed. Construct a 90% confidence interval for the difference µµ₂ between the mean recovery time when treated with

multivitamins (μ,) and the mean recovery time when treated with vitamin C only (H2). Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

What is the lower limit of the 90% confidence interval?

What is the upper limit of the 90% confidence interval?

Answer:

I assume the higher notes would make the rice vibrate more easily, so a flute.

Vab= E = 20V

because I = 0 and the voltage drop across the resistances R1 and R2 is also 0.

Second circuit:

Vab = 10V (no voltage drop across R1)

Vcd= E2-E1 = 20V

Series connection of voltage sources. But the sources are connected to the contrary and voltage drop across R1 or R2 is 0 V.

Answer:

98.11 I think

Explanation:

I really hope this helps have a wonderful day

Answer:

Hydrogen

1 valence electron

Helium

2 valence electrons

lithium

1 valence electrons

beryllium

2 valence electrons

boron

3 valence electrons

carbon

4 valence electrons

nitrogen

5 valence electrons

oxygen

6 valence electrons

flourine

7 valence electrons

neon

8 valence electrons

sodium

1 valence electron

magnesium

2 valence electrons

aluminum

3 valence electrons

silicon

4 valence electrons

phosphorus

5 valence electrons

Answer:  17 Chlorine -1, +1, (+2), +3, (+4), +5, +7

18 Argon 0

19 Potassium +1

20 Calcium +2

Explanation:

In order for the refrigerator not to tip over, the maximum acceleration of  1.86 m/s² must not be exceeded.

The term acceleration has to do with the rate at which velocity changes with time.

We have to take the moments at the tipping point of rotation as follows;

Clockwise moment = Anticlockwise moment

Hence;

F₂ * 1.58 m = F₁ * 0.67 m

The weight at half the width= 30 cm or 0.3 m

Height of refrigerator = 158 cm 0r 1.58 cm

Then;

m * a * 1.58 = m * 9.81 * 0.30

a = 1.86 m/s²

In order for the refrigerator not to tip over, the maximum acceleration of  1.86 m/s² must not be exceeded.

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Answer:

The power generated by the windmill is approximately 1.364 MW

Explanation:

The diameter of the windmill, d = 30 m

The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s

The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s

The pressure at the inlet, P₁ = 2 atm

The pressure at the outlet, P₂ = 1 atm

The density of air, ρ = 1.2 kg/m³

The power obtained from the windmill, 'P', is given as follows;

[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]

Where;

[tex]g_c[/tex] = 1.0 kg/(N·s²)

A = Cross-sectional rea of the the windmill =  π·D²/4 = π×(30 m)²/4 = 706.858347 m²

Plugging in the values, we get;

[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]

The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.

Answer:

I think it is transpiration

Answer:

transpiration is the right answer

The percent difference between two numbers [tex]x[/tex] and [tex]y[/tex] is given by

[tex]\dfrac{|y-x|}x \times 100\%[/tex]

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from [tex]x[/tex] to [tex]y[/tex] or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

[tex]\dfrac{|6-3|}3 \times 100\% = 100\%[/tex]

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

[tex]\dfrac{3-6}3\times100\%=-50\%[/tex]

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take [tex]x[/tex] from the left column and [tex]y[/tex] from the right column.

[tex]\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%[/tex]

[tex]\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%[/tex]

[tex]\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%[/tex]

The percentage difference are 2.82%, 4.17%, 4.17%.

The percent difference between two numbers a and b is given by formula:

[tex]\frac{y-x}{x}*100[/tex]

We are given absolute value as we are only concerned about the absolute percent difference.

Making comparisons to object measurements determines that the differences should be computed relative to object measurements.

Here take from the left column and  from the right column.

a) [tex]\frac{7.3-7.1}{7.1} *100=2.82[/tex]%

b) [tex]\frac{5.0-4.8}{4.8} *100=4.17[/tex]%

c) [tex]\frac{7.5-7.2}{7.2} *100=4.17[/tex]%

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The uniform acceleration of the car is 4.485 m/s².

The uniform acceleration of the car is calculated as follows;

v² = u² + 2as

a = (v² - u²)/2s

where;

a = (41² - 28²)/(2 x 100)

a = 4.485 m/s²

Thus, the uniform acceleration of the car is 4.485 m/s².

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Based on the calculations, the ratio of the intensity of Ir to Ip is equal to 50.12.

Mathematically, sound intensity level can be calculated by using this formula:

[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]

Where:

I is the intensity of the sound.

Making I the subject of formula, we have:

I = Io × [tex]10^{\frac{\beta }{10} }[/tex]

For Ip, we have:

Ip = Io × [tex]10^{\frac{104 }{10} }[/tex]

Ip = Io × [tex]10^{10.4}[/tex]

For Ir, we have:

Ir = Io × [tex]10^{\frac{121 }{10} }[/tex]

Ir = Io × [tex]10^{12.1}[/tex]

Now, we can find the ratio of the intensity:

Ir/Ip = Io × [tex]10^{12.1}[/tex]/Io × [tex]10^{10.4}[/tex]

Ir/Ip = [tex]10^{12.1}[/tex]/[tex]10^{10.4}[/tex]

Ir/Ip = [tex]10^{12.1 -10.4}[/tex]

Ir/Ip = [tex]10^{1.7}[/tex]

Ir/Ip = 50.12.

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Answer:

Reflection can simply be defined as the reflection of light when it strikes the medium on a plane. Refraction can be defined as the process of the shift of light when it passes through a medium leading to the bending of light. The light entering the medium returns to the same direction.

Answer:

reflection is your image and refraction is light

Answer:

[tex]\mathbf{V_d = 3.2 \times 10^5 \ m/s}[/tex]

Explanation:

[tex]\text{The speed of the protons can be estimated by using the formula:}[/tex]

[tex]V_d = \dfrac{I}{enA}[/tex]

[tex]where;[/tex]

[tex]\text{e = proton charge}[/tex]

[tex]\text{n = No. of protons per unit volume}[/tex]

[tex]\text{A = area of aperture}[/tex]

[tex]V_d = \dfrac{91 \times 10^{-9} \ A}{(1.602 \times 10^{-19} \ C (7.0 \times 10^6 \ m^{-3} ) (\pi) (0.284 \ m)^2}[/tex]

[tex]\mathbf{V_d = 3.2 \times 10^5 \ m/s}[/tex]

Answer:

friction help to slow motion in other word it oppose motion, but in a frictionless environment object would move with difficult stopping point.

Answer:

yes friction is needed hope this helps might of been to long tho

Answer:

This is something for biology, it shouldnt be in the physics subject

Explanation:

Answer:

a)  A' =  0.345  m,  b)  f = 2,800 Hz

Explanation:

b) The angular velocity of a simple harmonic motion is

        w =[tex]\sqrt{\frac{k}{m} }[/tex]

angular velocity and frequency are related

        w = 2π f

we substitute

        f = 1 /2π   √k/m

indicates that the initial frequency value f = 3.96 Hz

in this case the mass is reduced by half

       m ’= m / 2

we substitute

       f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]

       f = √1/2    (2π √k/m)

       f = 1 /√2  3.96

       f = 2,800 Hz

a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.

When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment

initial instant. Right before the division

        p₀ = (m₁ + m₁) v

final instant. Right after the split

        p_f = m₁ v '

        p₀ = p_f

        (2 m₁) v = m₁ v ’

        v ’= 2v

At this point we can use conservation of energy for the system with only half the block.

Starting point. Where the block divides

         Em₀o = K = ½ m v'²

Final point. Point of maximum elongation

          Em_f = Ke = ½ k A²

how energy is conserved

         Em₀ = Em_f

         ½ m’ v’² = ½ k A’²

we substitute the previous expressions

         ½ m/2 (2v)² = ½ k A’²

         A’² = 2  m v² / k                       (1)

Let's use the conservation of energy with the initial conditions, before dividing the block

          ½ m v2 = ½ k A2

          A² = mv² / k = 5.95 10⁻² m²

we substitute in 1

         A'² = 2 A²

          A ’²=  2 5.95 10⁻²

          A ’²= 11.9 10⁻² m

          A' =  0.345  m

Answer:

A

Explanation:

AWNSER:

awnser:

C

explanation:

Answer:

-3.77 m/s

3.85 m/s

Explanation:

given that

Frequency at stationary = 180 Hz

Beat frequency = 2 Hz

Using Doppler effect, we know that

f' = f[(v ± v0) / (v ± vs)], where

v = speed of sound, 343 m/s

v0 = speed of the observer, 0

vs = speed of light, ?

f = stationary frequency, 180 Hz

f' = stationary ± beat frequency, 180 ± 2

Applying the formula, we have

f' = f[(v ± v0) / (v ± vs)]

182 = 180 [(343 + 0) / (343 + vs)]

182/180 = 343 / 343 + vs

343 + vs = 343 * 180/182

343 + vs = 339.23

vs = 339.23 - 343

vs = -3.77 m/s

Again, using

f' = f[(v ± v0) / (v ± vs)]

178 = 180 [(343 + 0) / (343 + vs)]

178/180 = 343 / 343 + vs

343 + vs = 343 * 180/178

343+ vs = 346.85

vs = 346.85 - 343

vs = 3.85 m/s

Answer:

The final velocity of the baseball as it leaves the bat is 40 m/s

Explanation:

The given parameters of the baseball and bat are;

The mass of the baseball = 0.15 kg

The mass of the bat = 1.0 kg

The velocity of the ball before collision, v₁ = 40 m/s

The velocity of the bat before collision, v₂ = -40 m/s

The coefficient of restitution, e = 0.50

Let, 'v₃', and 'v₄' represent the final velocity of the ball and the bat respectively after collision, we have;

Taking the final velocity of the bat, v₄ = 0 m/s

According to Newton's Law of restitution

e = (v₃ - v₄)/(v₁ - v₂)

∴ 0.5 = (v₃ - v₄)/(40 - (-40))

80 × 0.5 = 40 = (v₃ - v₄)

v₃ - v₄ = 40

v₃ =  40 + v₄ = 40 + 0 = 40

The final velocity of the baseball as it leaves the bat, v₃ = 40 m/s.

Explanation:

this is the answer for your question. if you have any doubt.

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The current is 1.13* 10^{-4}A

Given that r = radius of the coil = 4 cm = 0.04 m

Area of coil is given as

A = πr²

A = (3.14) (0.04)² = 0.005024 m²

N = Number of turns = 500

R = Resistance = 600 Ω

B = magnetic field = (0.0120)t + (3 x 10⁻⁵) t⁴

Taking derivative at both the side

[tex] \frac{dB}{dt} = (0.120 + (12 \times 10^-5)t^3)[/tex]

Induced current is given as

[tex]i= (\frac{NA}{R} )( \frac{db}{dt} )[/tex]

[tex]i \: = (\frac{NA}{R})(12 \times 10 ^{-5} )t^3[/tex]

substituting the value t = 5

[tex]i = ( \frac{(500)(0.005024)}{600}) (12 \times 10 ^{-5} )5^3[/tex]

[tex]i = 1.13 \times 10 ^{ - 4} A[/tex]

Hence the current is

[tex]1.13 \times 10^-4A[/tex]

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The Electric current is 1.11* 10^{-4}A


Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

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