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3 years ago

How does proximity to oceans affect Temperatures and rainfall amounts?

Physics

1 answer:

vampirchik [111]3 years ago

4 0

Answer:

Because they absorb and emit heat more steadily than land, oceans help moderate temperatures in coastal areas, making winters warmer and summers cooler. Winds that blow in from across the oceans bring more rainfall and higher humidity.

Explanation:

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An organ pipe is 248 cm long.(a) Determine the fundamental overtone if the pipe is closed at one end.(b) Determine the first aud

Phantasy [73]

(a) 34.6 Hz

The fundamental frequency of a pipe closed at one end is given by

$f_1 = \frac{v}{4 L}$

where

v = 343 m/s is the speed of the sound in air

L is the length of the pipe

In this problem,

L = 248 cm = 2.48 m

So, the fundamental frequency is

$f_1 = \frac{343 m/s}{4 (2.48 m)}=34.6 Hz$

(b) 103.8 Hz

In a open-closed pipe, only odd harmonics are produced; therefore, the frequency of the first overtone (second harmonic) is given by:

$f_2 = 3 f_1$

where

$f_1 = 34.6 Hz$ is the fundamental frequency.

Substituting into the equation,

$f_2 = 3 (34.6 Hz)=103.8 Hz$

(c) 173 Hz

The frequency of the second overtone (third harmonic) is given by:

$f_3 = 5 f_1$

where

$f_1 = 34.6 Hz$ is the fundamental frequency.

Substituting into the equation,

$f_3 = 5 (34.6 Hz)=173 Hz$

(d) 242.2 Hz

The frequency of the third overtone (fourth harmonic) is given by:

$f_4 = 7 f_1$

where

$f_1 = 34.6 Hz$ is the fundamental frequency.

Substituting into the equation,

$f_4 = 7 (34.6 Hz)=242.2 Hz$

(e) 69.2 Hz

The fundamental frequency of a pipe open at both ends is given by

$f_1 = \frac{v}{2 L}$

where

v = 343 m/s is the speed of the sound in air

L is the length of the pipe

In this problem,

L = 248 cm = 2.48 m

So, the fundamental frequency is

$f_1 = \frac{343 m/s}{2 (2.48 m)}=69.2 Hz$

(f) 138.4 Hz

In a open-open pipe, both odd and even harmonics are produced; therefore, the frequency of the first overtone (second harmonic) is given by:

$f_2 = 2 f_1$

where

$f_1 = 69.2 Hz$ is the fundamental frequency.

Substituting into the equation,

$f_2 = 2 (69.2 Hz)=138.4 Hz$

(g) 207.6 Hz

The frequency of the second overtone (third harmonic) in an open-open pipe is given by:

$f_3 = 3 f_1$

where

$f_1 = 69.2 Hz$ is the fundamental frequency.

Substituting into the equation,

$f_3 = 3 (69.2 Hz)=207.6 Hz$

(h) 276.8 Hz

The frequency of the third overtone (fourth harmonic) in an open-open pipe is given by:

$f_4 = 4 f_1$

where

$f_1 = 69.2 Hz$ is the fundamental frequency.

Substituting into the equation,

$f_4 = 4 (69.2 Hz)=276.8 Hz$

7 0

3 years ago

Compare the Vf calculated at the point of impact to the horizontal velocity you calculated using Δx and Δy. Were the vf and the

34kurt

Answer: Velocities are not equal

Explanation: A projectile usually have a given initial velocity, the velocity must be resolve into components of x and y with respect to the velocity of the projectile in a certain angle. In a projectile trajectory the final velocity for the y- component is always 0m/s since the object reaches its maximum point.

4 0

3 years ago

A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. How mu

Nonamiya [84]

Complete Question:

A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. b) How much kinetic energy does it gain? The mass of a proton is 1.67262 × 10−27 kg, the mass of a neutron 1.67493 × 10−27 kg and the charge on an electron −1.60218 × 10−19 C. Answer in units of J\

b) what is the speed?

Answer:

a) the kinetic energy gained = 6.42 * 10⁻¹⁶ J

b) the speed of the particle, v = 619328.3 m/s

Explanation:

q = 1.602 *10⁻¹⁹C

V = 4.01 kV = 4.01 * 10³ V

Work done by the deuteron = qV

Work done by the deuteron = 1.602 * 10⁻¹⁹ * 4.01 *10³

Work done = 6.42 * 10⁻¹⁶ J

Kinetic Energy gained = work done

Kinetic Energy gained by the deuteron = 6.42 * 10⁻¹⁶ J

B) The formula for Kinetic Energy is given by:

KE = 1/2 Mv²

Let the mass of the proton be m₁ = 1.67262 × 10⁻²⁷kg

Let the mass of the neutron be m₂ = 1.67493 × 10−27 kg

M = m₁ + m₂

KE = 1/2 ( m₁ + m₂)v²

Let v = speed of the deuteron

From part (a)

KE = 6.42 * 10⁻¹⁶ J

1/2 ( m₁ + m₂)v²= 6.42 * 10⁻¹⁶

0.5 * (1.67262 + 1.6749) *10⁻²⁷ * v² = 6.42 * 10⁻¹⁶

v = 619328.3 m/s

6 0

3 years ago

A square loop of side length a =4.5 cm is placed a distance b = 1.1 cm from a long wire carrying a current that varies with time

sergey [27]

Answer:

a)Current will flow perpendicularly.

b)Magnitude of flux will be 2.987 N m2 C−1

3 0

3 years ago

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The slope of the line on any speed vs time graph is equal to the objects ?

vivado [14]

i think its a bar graph

i hope this works

3 0

3 years ago