PHYSICS COLLEGE

The New York Wheel is the world's largest Ferris wheel. It's 183 meters in diameter and rotates once every 37.3 min.1. Find the magnitude of the average velocity at the wheel's rim, over a 7.40-
min interval.

2.Find the magnitude of the average acceleration at the wheel's rim, over a 7.40-
min interval.

Answers

Answer 1

Answer:

Answer:

Velocity =0.241 m/s

Acceleration = 7.21e-4 m/s²

Explanation:

The wheel travels through

Θ = (7.40/37.3)*360º = 71.42º

and so the length of the line segment connecting the initial and final position is

L = 2*L*sin(Θ/2) = 2 * (183m/2) * sin(71.42º/2) = 107 m

so the average velocity is

v = L / t = 107m / 7.40*60s = 0.241 m/s

Initially, let's say the velocity is along the +x axis:

Vi = π * 183m / (37.3*60s) i = 0.257 m/s i

Later, it's rotated through 71.42º, so

Vf = 0.257m/s * (cos71.42º i + sin71.42º j) = [0.0819 i + 0.244 j] m/s

ΔV = Vf - Vi = [(0.0819 - 0.257) i + 0.244 j] m/s = [-0.175 i + 0.244 j] m/s

which has magnitude

|ΔV| = √(0.175² + 0.244²) m/s = 0.300 m/s

Then the average acceleration is

a_avg = |ΔV| / t = 0.300m/s / (7.40*60s) = 6.76e-4 m/s²

The instantaneous acceleration is centripetal: a = ω²r

a = (2π rads / (37.3*60s)² * 183m/2 = 7.21e-4 m/s²

Answer 2

Answer:

Answer:

$v = 0.24 m/s$

$a = 6.75 * 10^(-4) m/s^2$

Explanation:

Given that wheel completes one round in total time T = 37.3 min

so angular speed of the wheel is given as

$\omega = (2\pi)/(T)$

$\omega = (2\pi)/(37.3) rad/min$

now the angle turned by the wheel in time interval of t = 7.40 min

$\theta = \omega t$

$\theta = ((2\pi)/(37.3))(7.40) = 0.4\pi$

PART 1)

Now the average velocity is defined as the ratio of displacement and time

here displacement in given time interval is

$d = 2Rsin(\theta)/(2)$

R = radius = 91.5 m

$d = 183sin(0.2\pi) = 106.8 m$

Now time to turn the wheel is given as

$t = 7.40 min = 444 s$

now we have

$v = (d)/(t) = (106.8)/(444)$

$v = 0.24 m/s$

PART 2)

Now average acceleration is defined as ratio of change in velocity in given time interval

here velocity of a point on its rim is given as

$v = R\omega$

$v = (91.5)((2\pi)/(37.3* 60))$

$v = 0.257 m/s$

now change in velocity when wheel turned by the above mentioned angle is given as

$\Delta v = 2vsin(\theta)/(2)$

$\Delta v = 2(0.257)sin(0.2\pi)$

$\Delta v = 0.3 m/s$

time interval is given as

$t = 7.40 min = 444 s$

now average acceleration is given as

$a = (0.3)/(444)$

$a = 6.75 * 10^(-4) m/s^2$

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Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Answers

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

Answers

Answer:

The distance covered by the bird before feeding is $4.55 * 10^(5)$ m.

Explanation:

As the bird consumes 4 g of fat before flying, the amount of initial food energy ( $E_(F)$ ) stored by it is given by

$E_(F) = 4 g * 9.4 (food) cal = 37.6 (food) cal$

So the mechanical energy stored by the bird ( $E_(M)$ ) is given by

$E_(M) = E_(F) * 4186 J = 1.57 * 10^(5) J$

Given, the power consumed by the bird $P = 3.7 W$

So, the time ( $t$ ) required to consume this power by the bird is

$t = (E_(M))/(P) = (1.57 * 10^(5) J)/(3.7 W) = 4.24 * 10 ^(4) s$

As the bird flies at an average speed ( $v$ ) of $10.7 ms^(-1)$ , so the distance ( $d$ ) covered by the bird before feeding again is given by

$d = v * t = 10.7 ms^(-1) * 4.25 * 10 ^(4) s = 4.55 * * 10^(5) m$

The distance of the bird'sflight before him/her feeds again is mathematically given as

d = 4.55* 10^{5} m

What is the distance of the bird's flight before him/her feeds again?

Question Parameter(s):

a bird that flies at an averagespeed of 10.7 m/s

its body fat reserves at an average rate of 3.70 W

the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories,

Generally, the initial food energy is mathematically given as

Ex= 4 g*9.4

Ex= 37.6cal

Therefore, the mechanical energy

Em = Ex * 4186

Em = 1.57*10^{5} J

In conclusion, time of flight

$t = (E_(M))/(P) \n\n t=(1.57 *10^(5) J)/(3.7 W) \n$

t= 4.24*10 ^{4} s

Th distance hence is

d = v* t

d= 10.7 *4.25*10 ^{4}

d = 4.55* 10^{5} m

Answers

Answer:

I am sure it is A because no chemical change occurs and it is a physical change. If you can Brainllest than that would be great but if you wanna you don't have to. Hope this helps!! If wrong sorry.

Explanation:

Which correctly describes how the energy of a wave on the electromagnetic spectrum depends on wavelength and frequency?A.
Energy decreases with decreasing wavelength and decreasing frequency.
B.
Energy increases with decreasing wavelength and increasing frequency.
C.
Energy increases with decreasing wavelength and decreasing frequency.
D.
Energy decreases with increasing wavelength and increasing frequency.

Answers

Answer:

B. Energy increases with decreasing wavelength and increasing frequency.

Explanation:

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200000 people. Satellites observing these waves from space measured 800 from one wave crest to the next and a period between waves of 1.0 hour.Part AWhat was the speed of these waves in m/s?Express your answer using two significant figures.=Part BWhat was the speed of these waves in km/h ?Express your answer using two significant figures.=

Answers

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

$v=f \lambda$

Where:

$v$ is the speed

$f=(1)/(T)$ is the frequency, which has an inverse relation with the period $T=1 h$

$\lambda=800 km$ is the wavelength

Solving with the given units:

$v=(1)/(T)\lambda$

$v=(1)/(1 h)800 km$

$v=800.00 km/h$ This is the speed of the wave in km/h

Transforming this speed to m/s:

$v=800.00 (km)/(h) (1 h)/(3600 s) (1000 m)/(1 km)$

$v=222.22 m/s$ This is the speed of the wave in m/s

What is the magnitude of the force you must exert on the rope in order to accelerate upward at 1.4 m/s2 , assuming your inertia is 63 kg ? Express your answer with the appropriate units.

Answers

Answer:

The magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Explanation:

The magnitude of force, you must exert can be estimated as follows;

Since it is upward motion, we must consider acceleration due to gravity which opposes the upward motion.

F = m(a+g)

where;

F is the magnitude of the upward force

m is your mass, which is the measure of inertia = 63kg

a is the acceleration of the rope = 1.4 m/s²

F = 63(1.4 + 9.8)

F = 63(11.2)

F = 705.6 N

Therefore, the magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Answer:

705.6 N

Explanation:

Force: This can be defined as the product of mass a acceleration.

The S.I unit of force is Newton.

The expression for the force on the rope in order to accelerate upward is given as,

F-W = ma .......................... Equation 1

Where F = Force exerted on the rope, W = weight of the rope, m = mass of the rope, a = acceleration.

But,

W = mg........................ Equation 2

Where g = acceleration due to gravity

substitute equation 2 into equation 1

F-mg = ma

F = ma+mg

F = m(a+g).............. Equation 3

Given: m = 63 kg, a = 1.4 m/s²

Constant: g = 9.8 m/s²

Substitute into equation 3

F = 63(1.4+9.8)

F = 63(11.2)

F = 705.6 N

The magnitude of the force exerted on the rope = 705.6 N

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