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4 years ago

To determine a waves' frequency, you must know the

1 answer:

6 0

The correct answer is<span> number of oscillations in a given period of time

This is measured in what is called the Hertz measurement and the period of time is usually taken to be per second.</span>

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A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang

Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ = density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0

3 years ago

15 POINTS!!!! PLEASE ANSWER IM DOING AN ESSAY AND I ONLY HAVE 10 MINUST LEFT TO SEND IT IN!!!!

Colt1911 [192]

Answer:

Carbon dioxide is added to atmosphere naturally when organisms respire or decompose

Or through human activities, such as the burning of fossil fuels and forests and the production of cement.

3 0

3 years ago

Explain how the plants of a tropical rain forest contribute to the characteristic hot, humid weather associated with the area. U

Talja [164]

Answer: The atmosphere of trophic rainforest is hot and humid due to high rate of transpiration.

Explanation:

The tropical rainforest is a biome which exhibit rich biodiversity of plants and animals. The average temperatures in this region remain high with warm summer. It remains frost free. The soil is nutrient deficient. Due to hot temperature the rate of transpiration remains high as a result the concentration of the water vapors remain high. This is responsible for increasing the humidity in the atmosphere also this region receives appreciable amount of rainfall annually. The average rainfall range is 200-450 centimeters.

6 0

3 years ago

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago

In a longitudinal wave, the places where particles of matter are closer together called

Blizzard [7]

Explanation:

I hesitate to articulate upon the highest degree of accuracy!

5 0

4 years ago