Which is NOT involved in photosynthesis

A long hollow cylindrical conductor (inner radius = 2.0 mm, outer radius = 4.0 mm) carries a current of 12 a distributed uniform

11Alexandr11 [23.1K]

Here we can use ampere'a law to find the magnetic field

$\int B.dl = \mu_o i_{en}$

$B*2 \pi r = \mu_o (i_1 + \frac{i_2*\pi(3^2 - 2^2)}{\pi(4^2 - 2^2}$

$B*2 \pi*0.003 = 4\pi * 10^{-7} (12 + 5)$

$B = 4\pi * 10^{-7} (12 + 5)$

$B = 1.13 * 10^{-3} T$

3 0

3 years ago

Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as object A t

Yuki888 [10]

Answer:

1:4

Explanation:

We have, two identical objects A and B fall from rest from different heights to the ground.

Object B takes twice as long as object A to reach the ground. It is required to find the ratio of the heights from which A and B fell. Let $h_A\ \text{and}\ h_B$ are the height for A and B respectively. So,

$\dfrac{h_A}{h_B}=\dfrac{(1/2)gt_A^2}{(1/2)gt_B^2}\\\\\dfrac{h_A}{h_B}=\dfrac{t_A^2}{t_B^2}$

We have,

$t_B=2t_A$

So,

$\dfrac{h_A}{h_B}=\dfrac{t_A^2}{(2t_B)^2}\\\\\dfrac{h_A}{h_B}=\dfrac{1}{4}$

So, the ratio of the heights from which A and B fell is 1:4.

7 0

3 years ago

what is the potential difference across the headlight bulbs when the starter motor is operated, requiring an additional 39 a fro

frosja888 [35]

The starter motor's potential difference across the headlight bulbs is 38.45V, requiring an additional 39 a from the battery. Voltage, also known as potential difference.

It is sometimes described as the amount of work needed to move a test charge between two sites, expressed as a unit of charge. Volt is the potential difference's SI unit (V). We only take into account the charge between the locations P and Q when current moves between them in an electric circuit. Electric potential difference between two sites is referred to as voltage, also known as electric pressure, electric tension, or (electric) potential difference. an electric field that is static.

Vh = I*Rn

Vh = 39/5.476*5.40v

Vh = 38.45v

Learn more about voltage here

brainly.com/question/13521443

#SPJ4

5 0

1 year ago

100%

xxMikexx [17]

Answer:

1. The elastic potential energy is 0.0176 Joules

2. The kinetic energy of the pinball the instant it leaves the spring is 0.0176 Joules

3. The speed of the pinball the instant it leaves the spring is approximately 2.42212 m/s

4. The height of the part where the pinball is located on the machine above the ground is approximately 0.213 meters

Explanation:

The spring constant of the pinball machine's plunger, k = 22 N/m

The amount by which the pinball machine's plunger is compressed, x = 0.04 m

The mass of the pinball ball, m = 0.006 kg

1. The elastic potential energy, P.E. = 1/2·k·x²

By substitution, we get;

P.E. = 1/2 × 22 N/m × (0.04 m)² = 0.0176 J

The elastic potential energy, P.E. = 0.0176 J

2. At the instant the pinball leaves the spring, the plunger and therefore the force of the plunger no longer acts on the pinball

Since there are no external forces acting on the pinball to increase the speed of the pinball after it leaves the spring, the velocity reached is its maximum velocity, and therefore, the kinetic energy, K.E. is the maximum kinetic energy which by the conservation of energy, is equal to the initial potential energy

Therefore;

K.E. = P.E. = 0.0176 J

The kinetic energy of the pinball the instant it leaves the spring, K.E.= 0.0176 J

3. The kinetic energy, K.E., is given by the following formula;

K.E. = 1/2·m·v²

Where;

v = The speed or velocity of the object having kinetic energy K.E.

Therefore, from K.E. = 0.0176 J, and by plugging in the values of the variables, we have;

K.E. = 0.0176 J = 1/2 × 0.006 kg × v²

v² = 0.0176 J/(1/2 × 0.006 kg) = 88/15 m²/s²

v = √(88/15 m²/s²) ≈ (2·√330)/15 m/s ≈ 2.42212 m/s

The speed of the pinball the instant it leaves the spring, v ≈ 2.42212 m/s

4. The height of the pinball is given by the following kinematic equation of motion;

$v_h$ ² = u² - 2·g·h

Where;

$v_h$ = The velocity of the pinball at the given height = 1.3 m/s

u = v ≈ 2.42212 m/s (The initial velocity of the pinball as it the spring)

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height of the pinball above the ground

We get;

$v_h$ ² = 1.3² = 2.42212² - 2 × 9.8 × h

∴ h = (2.42212² - 1.3²)/(2 × 9.8) ≈ 0.213

The height of the part where the pinball is located on the machine above the ground, h ≈ 0.213 m

5 0

3 years ago

A 2640-Hz sound source is moving at 15.0m/s toward a stationary observer. What is the frequency heard by the observer if the spe

Dvinal [7]

Answer:

The frequency heard by the observer is 2760.73 hertz. Explanation:

Frequency of source, f = 2640 Hz

Velocity of source, $v_s=15\ m/s$

The speed of sound, v = 343 m/s

Let f' is the frequency heard by the observer. According to Doppler's effect, the frequency of the observer is given by :

$f'=\dfrac{fv}{v-v_s}$ (as the source is moving towards observer)

$f'=\dfrac{2640\times 343}{343-15}$

f' = 2760.73 Hz

So, the frequency heard by the observer is 2760.73 hertz. Hence, this is the required solution.

8 0

4 years ago