PHYSICS COLLEGE

The height of the upper falls at Yellowstone Falls is 33 m. When the water reaches the bottom of the falls, its speed is 26 m/s. Neglecting air resistance, what is the speed of the water at the top of the falls?

Answers

Answer 1

Answer:

Answer:

Speed of water at the top of fall = 5.40 m/s

Explanation:

We have equation of motion

$v^2=u^2+2as$

Here final velocity, v = 26 m/s

a = acceleration due to gravity

$a=9.8m/s^2 \n$

displacement, s = 33 m

Substituting

$26^2=u^2+2* 9.8 * 33\n\nu^2=29.2\n\nu=5.40m/s \n$

Speed of water at the top of fall = 5.40 m/s

Related Questions

In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant
A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3
A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?
Mudflows composed of soil, volcanic debris, and water can occur as the result of an explosive volcanic eruption. What are these mudflows called?
What is matter? explain and give example

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of the car applies a torque of 456 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

Answers

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x $F_(s)$ = τ1

$F_(s)$ = the static frictional force (N)

$F_(s)$ = τ1 /r

= 456 N/0.380 m

= 1200 N

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:

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:

$v = 0.24 m/s$

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

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

$a = (0.3)/(444)$

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

An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe?

Answers

Answer:

The fraction of its volume inside liquid is increased .

Explanation:

According to principle pf floatation , an object floats on the surface of water

when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .

In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .

Which of these factors make hydrogen fuel cells a better option than burning fossil fuels? A.Hydrogen fuel cells have a higher energy efficiency. B.Hydrogen fuel cells create less pollution. C.Burning fossil fuels relies on outdated devices and technology. D.Hydrogen is the most abundant element in the universe. E.Hydrogen fuel cells are more expensive than fossil fuels.

Answers

Answer: i think it is B

Explanation:

Which one A, B, C, D or the last one?

Answers

Answer:

Explanation:

Answer is C.

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$