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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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brilliants [131]

Answer:

d

Explanation:

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A bicyclist is initially traveling at 3 m/s. The bicyclist accelerates at 1 m/s2 for 5 seconds.

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The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

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2 years ago

A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the

dsp73

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

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2 years ago

Power is equivalent to the energy released divided by the time taken to release the energy. true false

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2 years ago

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

$O=(0,0)$

$A=(0,-b)$

$B=(h,0)$

We need to calculate the position vector of AB

$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

$u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}$

We need to calculate the force acting along the edge

$\hat{F}=F(u_{AB})$

$\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})$

We need to calculate the net moment

$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0

3 years ago