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2 years ago

8

Why are many mitochondria needed in cells that move, like muscle cells?

1 answer:

Deffense [45]2 years ago

3 0

Muscle cells have to use a lot of energy and the Mitochondria is known as the "power house" of the cell... I would say the number of mitochondria needed is increased because more power and energy is being used in that cell than a normal one (idk if this helps or if that's the answer you were looking for)

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Which is not a device for reproducing sound? phonograph MP3 player gramophone microphone

m_a_m_a [10]

Answer:

microphone

Explanation:

3 0

3 years ago

What is the name of NASA's special plane?

saveliy_v [14]

Answer:

Boeing

if im wrong sue me ¯\_(ツ)_/¯

8 0

3 years ago

A ball is dropped from rest at point O . It passes a window with height 3.8 m in time interval tAB = 0.02 s.Identify the correct

Taya2010 [7]

Answer:

VB − VA = g tAB & (VA + VB)/2 = h / tAB

Explanation:

s = h = Displacement

tAB = t = Time taken

VA = u = Initial velocity

VB = v = Final velocity

a = g = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{3.8-\frac{1}{2}\times 9.8\times 0.02^2}{0.02}\\\Rightarrow u=189.902\ m/s$

$v=u+at\\\Rightarrow v=189.902+9.8\times 0.02\\\Rightarrow v=190.098\ m/s$

$\frac{v+u}{2}=\frac{190.098+189.902}{2}=190\ s$

$\frac{h}{t}=\frac{3.8}{0.02}=190\ s$

Hence, the equations VB − VA = g tAB & (VA + VB)/2 = h / tAB will be used

3 0

3 years ago

A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a

marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$ .

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$ .

The initial velocity:

$v_0 = - 20 \frac{m}{s}$ ,

take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$ .

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$ .

$y(6 \ s) = \ 1000 m \ - 300 m$ .

$y(6 \ s) = \ 700 m$ .

So this its the height of the ball 6 seconds after being thrown.

6 0

3 years ago

A star produces 2x10^26 watts. how much energy does it lose every minutes

Step2247 [10]

Answer:

Energy loss per minute will be $120\times 10^{26}j$

Explanation:

We have given the star produces power of $2\times 10^{26}W$

We know that 1 W = 1 J/sec

So $2\times 10^{26}W=2\times 10^{26}J/sec$

Given time = 1 minute = 60 sec

So the energy loss per minute $=2\times 10^{26}\times 60=120\times 10^{26}j$

We multiply with 60 we have to calculate energy loss per minute

7 0

3 years ago