a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?​

Answer: B

Explanation:

Answers A and C are examples of physical weathering while B is chemical weathering when water and lime mix it creates a reaction

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²  

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²  

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

Answer:

T = 2.5 s

Explanation:

Given that,

Number of complete orbits = 10

Time, t = 25 seconds

We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.

So,

[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]

So, the period of the orbiting ball is equal to 2.5 seconds.

1. third law

2. first law

3. third law

4. second law

Answer:

The correct answer is -  91.4 nm

Explanation:

According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:

(h×c)/λ=13.6eV

here,

h - Planck constant

c - the speed of light

λ - wavelength

Placing the value in the formula for the wavelength

(6.626×10^−34J.s × 3×10^8 m/s)/λ  =  13.6 ×1.6 × 10^−19 J

λ≈91.4nm

Thus, the correct answer would be = 91.4 nm

Answer:

14 μm

Explanation:

The magnetic field due to a long straight wire is B = μ₀i/2πr where  μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current = 7.0 A and r = distance of credit card from magnetic field.

So r = μ₀i/2πB since B = 1000 gauss = 1000 G × 1 T/10000 G  = 0.1 T

r = 4π × 10⁻⁷ H/m × 7.0 A/(2π × 0.1 T)

r = 2 × 10⁻⁷ H/m × 7.0 A/0.1 T

r = 14 × 10⁻⁷ H/m × A/0.1 T

r = 140 × 10⁻⁷ m

r = 1.4 × 10⁻⁵ m

r = 14 × 10⁻⁶ m

r = 14 μm

Answer:

1) Magnesium metal/magnesium ion

2) Aluminum/aluminum ion

3)  Copper metal/copper(l) ion

Explanation:

The activity series is a series that shows the ease of reactivity of substances in an electrochemical cell.

The substances that are higher up in the series are more reactive in electrochemical cells.

Magnesium is the first element in the series that has the most negative redox potential  then followed aluminium.

Hence, according to Nernst,

1) Magnesium metal/magnesium ion

2) Aluminum/aluminum ion

3)  Copper metal/copper(l) ion

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Answer:

B motor

Explanation:

Answer:

3.17333333333? I hope I get it right

Explanation:

..................hello

Answer:

hey but the person at the top is right

Answer:

0.185

Explanation:

Volume of water displaced = 0.000250 ( volume of cup )

Mass of water displaced by cup = density of water X volume of water displaced

= 1000 X 0.000250 = 0.250 kg

Mass of water displaced = mass of cup + mass of pennies ( law of flotation)

0.25 = 0.0650 + mass of pennies

Mass of pennies = 0.2500 - 0.0650

= 0.185 kg

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Answer:

0.03 A

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

From ohm's law, the voltage, current and resistor are related by the following formula:

Voltage = current × resistor

V = IR

With the above formula, we can obtain the current in the circuit as follow:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

V = IR

12 = I × 470

Divide both side by 470

I = 12 / 470

I = 0.03 A

Thus, the current in the circuit is 0.03 A

Answer:

0.03 A

Explanation:

Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

0.03 A

Answer:

The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.

Explanation:

mark me brainlest

Answer:

The predicted height is 2.809 meters, writing this in centimeters we get (1m = 100cm):

h = 2.809 m = (2.809)*(100cm) = 280.9 cm

And the total energy is:

E = 6.696 J

Explanation:

First let's see the problem.

We have an object of mass m = 274g which is thrown upwards with an initial velocity v0 = 6.991 m/s, in a place with a gravitational acceleration of g = 8.7 m/s^2

When the object is on the air, the only force acting on it will be the gravitational force, then the acceleration of the object will be equal to the gravitational acceleration, then we can write:

a(t) = -8.7 m/s^2

Where the negative sign is because this acceleration points down.

Now to get the velocity of the object we can integrate over time to get:

v(t) = (-8.7 m/s^2)*t + v0

Where v0 is a constant of integration, which is the initial velocity, then we can write this as:

v(t) = (-8.7 m/s^2)*t + 6.991 m/s

Now we can integrate again over the time to get the position equation.

p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t + p0

Where p0 is the initial position, because the ball is being thrown from the ground, the initial position is 0.

Then the position equation is:

p(t) = (1/2)*(-8.7 m/s^2)*t^2 + (6.991 m/s)*t

Ok, now we know all the movement equations for the object.

The first thing we want to know is the maximum height of the object.

We know that the object reaches its maximum height when the velocity is zero (this is, the velocity stops being positive, meaning that the object stops going up, then in that time we have the maximum height)

We need to solve:

v(t) = 0m/s = (-8.7 m/s^2)*t + 6.991 m/s

(8.7 m/s^2)*t =  6.991 m/s

t =  6.991 m/s/( (8.7 m/s^2)  = 0.804 seconds

The maximum height of the object is given by:

p(0.804s) = (1/2)*(-8.7 m/s^2)*(0.804)^2 + (6.991 m/s)*(0.804) = 2.809 m

The maximum height of the object is 2.809 meters.

Now let's find the maximum energy.

Remember that the energy of an object can be written as the sum of the potential energy U and the kinetic energy K.

E = K + U

Such that for an object of mass m and velocity v, the kinetic energy is:

K = (1/2)*m*v^2

And for an object of mass m, at a height h from the ground and with gravitational acceleration g, the potential energy is:

U = m*g*h

Now, when the object is at its maximum height, the velocity is zero.

Then K = 0

And for conservation of energy, the total energy of the object becomes potential energy.

E = 0 + U

E = U

So if we find the potential energy at the maximum height of the object's path, we can find the total energy of the object.

We know that:

mass = m = 274g = 0.274 kg  (here i used that 1kg = 1000g)

height = h = 2.809 meters.

gravitational acceleration = g = 8.7 m/s^2

Then the potential energy at this point is:

U =  0.274 kg*(2.809 meters)*(8.7 m/s^2) = 6.696 J

This means that the total energy of the object is:

E = 6.696 J

Answer:

Explanation:

V = J/C

V = 20/1

= 20 v

Option A is the correct answer

Answer:

55.897905

Explanation:

1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth

9.80665 times 5.7=55.897905

Brainliest?

Answer:

60

Explanation:

Work Done= Force×Displacement in the direction of the force

W.D. = 10×6

W.D. = 10×0.6

W.D. = 6m

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

Answer:

The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy

Answer:

The rocket will appear larger than it actually is

Answer:

the distance is 6.46 cm.

Explanation:

Given

current in the wire, I = 4.2 A

magnitude of the magnetic field, B = 1.3 x 10⁻⁵ T

The distance from the wire is determined by using Biot-Savart Law;

[tex]B = \frac{\mu_o I}{2\pi r} \\\\r = \frac{\mu_o I}{2\pi B}[/tex]

Where;

r is the distance from the wire where the magnetic field is experienced

[tex]r = \frac{\mu_o I}{2\pi B}\\\\r = \frac{4\pi \times 10^{-7} \times 4.2 }{2\pi \times 1.3 \times 10^{-5}}\\\\r = 0.0646 \ m\\\\r = 6.46 \ cm[/tex]

Therefore, the distance is 6.46 cm.

Answer:

The de Broglie wavelength will remain the same because it does not depend on temperature.

Explanation:

de Broglie wavelength of a particle is independent of the temperature and hence the properties of emitted particle such as photoelectric effect, radioactive radiation etc. does not depend on the temperature.

Also, until unless the kinetic energy of a moving particle is not driven by the

thermal energy, the de Broglie wavelength  is independent of the temperature

Answer: 14.5 kg.m/s

Explanation:

Given

mass of baseball is [tex]m=0.153\ kg[/tex]

The initial speed of the ball is [tex]u=-44.5\ m/s[/tex]

the final speed of the ball is [tex]v=50.5\ m/s[/tex]

Impulse is given as a change in the momentum

[tex]\vec{J}=\Delta \vec{P}[/tex]

[tex]J=m(v-u)\\J=0.153(50.5-(44.5))\\J=0.153\times 95=14.535\ kg.m/s[/tex]

Change in momentum up to 3 significant figures is 14.5 kg.m/s

Impulse applied by a bat is also the same as the change in momentum

Answer:

the total momentum of the system before collision is 6.94 kgm/s

Explanation:

Given;

mass of the first cart, m₁ = 2.0 kg

mass of the second cart, m₂ = 1.8 kg

velocity of the first cart before collision, u₁ = 5.9 m/s

velocity of the second cart before collision, u₂ = -2.7 m/s

The total momentum of the system before collision is calculated as follows;

[tex]P_t = P_1 + P_2 \\\\P_t = m_1u_1 + m_2u_2\\\\P_t = (2\times 5.9) + (1.8 \times -2.7)\\\\P_t = 11.8 - 4.86\\\\P_t = 6.94 \ kgm/s[/tex]

Therefore, the total momentum of the system before collision is 6.94 kgm/s