PHYSICS HIGH SCHOOL

A wave travels at a constant speed. How does the frequency change if the wavelength increases by a factor of 2?A.The frequency decreases by a factor of 2. B.The frequency increases by a factor of 4. C.The frequency increases by a factor of 2. D.The frequency does not change.

Answers

Answer 1

Answer:

Answer : The frequency decreases by a factor of 2.

Explanation :

Given that the wave travels at a constant speed. The speed of the wave is given as :

$v=\nu* \lambda$

Where

υ is the frequency of the wave

and λ is the wavelength of the wave.

In this case, the speed is constant. So, the relation between the frequency and the wavelength is inverse.

$\nu\propto (1)/(\lambda)$

If the wavelength increases by a factor of 2, its frequency will decrease by a factor of 2.

Hence, the correct option is (A) " The frequency decreases by a factor of 2 ".

Answer 2

Answer: A.The Frequency decreases by a factor of 2.

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Answers

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In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running over the edge of the cliff above Ohio's Cuyahoga River in (Figure 1) , which is confined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft (≈ 6.7 m) across while falling 20 ft (≈ 6.1 m).What is the minimum speed with which he’d need to run off the edge of the cliff to make it safely to the far side of the river?

Express your answer to two significant figures and include the appropriate units.

Answers

The minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Further explanation:

As Captain Sam Brady jumps from the cliff, he moves in two dimension under the action of gravity.

Given:

The height of free fall of the captain Brady is $20\text{ ft}$ or $6.1\text{ m}$ .

The horizontal distance moved by the captain Brady is $22\text{ ft}$ or $6.7\text{ m}$ .

Concept:

The time required to free fall of a body can be calculated by using the expression given below.

$\left( { - s}\right)=ut-(1)/(2)g{t^2}$ ……. (1)

The displacement is considered negative because the captain is moving in vertically downward direction.

Here, $s$ is the distance covered by the body in free fall, $u$ is the initial velocity of the object, $g$ is the acceleration due to gravity and $t$ is the time taken in free fall of a body.

As the Caption jumps off the cliff, he has his velocity in the horizontal direction. The velocity of the captain in vertical direction is zero.

Substitute $0$ for $u$ in the equation (1) .

$s=(1)/(2)g{t^2}$

Rearrange the above expression for $t$ .

$\boxed{t=\sqrt {\frac{{2s}}{g}}}$ …… (2)

Converting acceleration due to gravity in $\text{ft}/\text{s}^2$ .

$\begin{aligned}g&=\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)\left( {\frac{{1.0\,{\text{ft/}}{{\text{s}}^{\text{2}}}}}{{0.305\,{\text{m/}}{{\text{s}}^{\text{2}}}}}} \right) \n&=32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}} \n \end{aligned}$

Substitute $20\text{ ft}$ for $s$ and $32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}$ for $g$ in equation (2) .

$\begin{aligned}t&=\sqrt {\frac{{2\left( {20\,{\text{ft}}} \right)}}{{\left( {32.16\,{\text{ft/}}{{\text{s}}^{\text{2}}}} \right)}}} \n&=1.116\,{\text{s}} \n \end{aligned}$

Therefore, the time taken by captain to free fall a height $20\text{ ft}$ is $1.116\text{ s}$ .

In the same time interval captain has to move $22\text{ ft}$ in horizontal direction. The acceleration is zero in horizontal direction. So, the velocity will be constant throughout the motion in the horizontal direction.

The distance travelled by captain in the horizontal direction is given by,

$x=v\cdot t$

Rearrange the above expression for $v$ .

$\boxed{v=(x)/(t)}$ …… (3)

Here, $x$ is the distance travelled in horizontal direction, $v$ is the velocity of the captain and $t$ is the time.

Substitute $22\text{ ft}$ for $x$ and $1.116\text{ s}$ for $t$ in equation (3) .

$\begin{aligned}v&=\frac{{22\,{\text{ft}}}}{{1.116\,{\text{s}}}} \n&=19.71\,{\text{ft/s}} \n \end{aligned}$

Thus, the minimum speed with which the captain Sam Brady of the US continental army had to run off the edge of the cliff to make it safely to the far side of the river is $\boxed{19.667\text{ ft/s}}$ or $\boxed{5.998\text{ m/s}}$ or $\boxed{6\text{ m/s}}$ or $\boxed{599.8\text{ cm/s}}$ .

Learn more:

1. Energy density stored in capacitor brainly.com/question/9617400

2. Kinetic energy of the electrons brainly.com/question/9059731

3. Force applied by the car on truck brainly.com/question/2235246

Keywords:

Free fall, projectile, gravity, 1780, Brady’s, leap, Captain, Sam Brady, US, continental army, enemies, Ohio’s, Cuyahoga river, 22 ft, 6.7 m, 20 ft, 6.1 m, minimum speed, run off, edge, cliff, safely, far side, river, 19.71 ft/s, 6 m/s, 6 meter/s, 5.99 m/s, 599.8 cm/s.

Final answer:

Using the principles of projectile motion from Physics, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to reach the far side of the river.

Explanation:

This problem can be solved using basic Physics, specifically projectile motion. Here, Captain Sam Brady had to run off the edge of the cliff to make it safely to the far side of the river which is 22 ft away while falling 20 ft down. We assume that he jumps horizontally (i.e., his initial vertical velocity is 0).

Firstly, we calculate the time for the vertical fall. Using the equation t = sqrt (2h/g) where h is height and g is the acceleration due to gravity (32.2 ft/s²), we get time t ≈ 1.12s (rounded to two significant figures).

Next, we can use this time to figure out his initial horizontal velocity needed. The equation v = d/t where v is velocity, d is distance, and t is time gives us v ≈ 19.64 ft/s (rounded to two significant figures).

So, Captain Sam Brady would need to run with an initial horizontal speed of approximately 19.64 ft/s to make it safely across the river.

Learn more about Projectile Motion here:

brainly.com/question/20627626

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A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest.(a from the instant the plane touches the runaway, what is the minimum time needed before it can come to a rest?
(b Can this plane land on a runaway that is only 0.800 km long?
shown work pls will reward alot of points

Answers

Answer:

a) t = 20 s, b) x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance

Explanation:

This is a kinematics exercise

a) in minimum time to stop,

v = vo + at

v = 0

t = -v0 / a

we calculate

t = -100 / (5.00)

t = 20 s

b) Let's find the length you need to stop

v² = vo² + 2 a x

x = -v0 ^ 2 / 2a

x = - 100² / 2 (-5.00)

x = 1000 m

As the runway is only 800 m long, the plane cannot land at this distance.

The amount of matter in a substance or object is called:
??? What the answer

Answers

Answer:

mass!!

Explanation:

Which set of terms best defines what affects kinetic energy to potential energy, respectively?

Answers

Velocity and height are the terms that best describes what affects kinetic energy and potential energy.

By definition, the kinetic energy is given by:

Where,

m: body mass

v: body speed

On the other hand, the potential energy is:

Where,

m: body mass

g: acceleration of gravity

h: height of the object

Therefore, a set of terms that affects kinetic and potential energy are speed and height, respectively.

Answer:

The set of terms is speed and height, respectively.

The acceleration due to gravity on the moon is about one-sixth its value on earth. If a baseball reaches a height of 67 m when wurth (jaw6696) – Ch 2 LT 3/4 – kreft – (61920) 2 thrown upward by someone on the earth, what height would it reach when thrown in the same way on the surface of the moon? Answer in units of m.