Acceleration and Velocity

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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Answer:

98.11 I think

Explanation:

I really hope this helps have a wonderful day

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.

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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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:

A

Explanation:

AWNSER:

awnser:

C

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: True

hope this helps!

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:

Answer:

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

Explanation:

Answer:

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

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.

you can send your doubt to:6369514784(what's app)

Answer:

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

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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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:

10,378.82 m/s is the second cosmic speed.

69,801 km is the radius of the synchronous orbit of a satellite.

Given

Mass of planet = 4.74 × [tex]10^{24}[/tex] kg

Radius of planet = 5870 km = 5870000m

First Cosmic speed = 7.34 km/sec

1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.

It can be calculated by

v = √2GM/r where,

v= Escape velocity of the satellite

G = Gravitational constant

M = Mass of planet

r = Radius of planet

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

2) Speed of the satellite at the given period

v = 2πr/T where,

T= Time period of rotation = 16.6 × 3600 seconds

r = v×T/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Hence

The Second Cosmic Speed i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.

And the radius of the synchronous orbit of a satellite is 69,801 km.

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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

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 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:

(A) By reducing friction

Answer:

A

Explanation:

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:

Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.

Accuracy means the state of being accurate, without any mistakes and the results should be 100% true.

Whereas, precision means, approximately true or almost true.

Hope it help you

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:

The metalloids are located on the right side of the periodic table in a "step-like" arrangement.

All of the possible metalloids are:

boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)

C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.

Kinetic energy is the energy possessed by a body due to its motion.

K.E = ¹/₂mv²

where;

The kinetic energy of a body is directly proportional to the mass of the object.

Thus, the correct relationship between kinetic energy and mass is Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.

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(a) The work done in moving the crate along the floor is 7,050 J.

(b) The work done in moving the crate vertically is 0 J.

W = Fdcosθ

where;

W = (1100 + 310) x 5

W = 7,050 J

W  = Fd cos(90)

W = 0

Thus, the work done in moving the crate along the floor is 7,050 J.

the work done in moving the crate vertically is 0 J.

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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]