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3 years ago

10

Which organelle is the powerhouse of the cell, the site of cellular respiration? A) 2 - nucleus B) 5 - endoplasmic reticulum C)

6 - Golgi bodies D) 9 - mitochondria

1 answer:

Paladinen [302]3 years ago

7 0

D) 9- mitochondria
Hope this helps :D

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Jobisdone [24]

Answer:

C) For every action, there is an equal and opposite reaction

Explanation:

Newton's Third Law

For every force, there is an equal and opposite reaction time. Explains why forces act in pairs.

Brainliest?

3 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

3 years ago

If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtime with a frequency of 1,056 Hz this over

Romashka-Z-Leto [24]

I have a hunch that you're not talking about overtime
OR overtune. I think you're going for overtone .

For a fundamental frequency of 528 Hz,
1,056 Hz is the second harmonic.

5 0

3 years ago

Ch 31 HW Exercise 31.10 7 of 15 Constants You want the current amplitude through a inductor with an inductance of 4.90 mH (part

sergey [27]

Answer:

f = 130 Khz

Explanation:

In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.

For an inductor, this relationship can be expressed as follows:

VL = IL * XL (1) , which is a generalized form of Ohm's Law.

XL is called the inductive reactance, and is defined as follows:

XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and L is the value of the inductance, in H.

Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:

f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz

5 0

3 years ago

A worker’s success in contributions to the common good is?

Zinaida [17]

Answer:

measured by how well she or he does the job

Hope this helps! （づ￣3￣）づ╭❤～

6 0

4 years ago