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2 years ago

Producer: (plants) (produces its own food) 1. The common name of my organism is: ______________

Physics

2 answers:

Lelu [443]2 years ago

6 0

1. Root
2. Food

This is the answer hope u like it and plz mark me as brainlyist.

Mama L [17]2 years ago

4 0

Answer: 1. Plant

2. Glucose

Explanation:

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Which is an example of a solution?

vladimir1956 [14]

Answer:

Explanation:

plato

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3 years ago

A catcher "gives" with a baseball when catching it. If the baseball exerts a force of 437 N on the glove, so that the glove is d

sergiy2304 [10]

437x9
is ur answer. I'm not sure tho hope it helps

5 0

3 years ago

A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

nalin [4]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

4 0

3 years ago

Read 2 more answers

In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of

Airida [17]

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

3 0

3 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$