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3 years ago

What is a chloroplast in a animal cell

Physics

2 answers:

Kay [80]3 years ago

6 0

An animal doesnt have a chloroplast

goldenfox [79]3 years ago

4 0

Only plant cells have chloroplast. It helps a plant do photosynthesis.

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A body of mass 3.0Kg is acted upon by a force of 24N, if the frictional force on the body is 13N.Calculate the acceleration of t

lisabon 2012 [21]

Answer:

Fnet=ma

24-13=3a

11/3=a

a=3.6m/s2

4 0

3 years ago

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago

Acceleration is rate of change of <br>A-Position<br>B-Time<br>C-Velocity<br>D-Speed

Drupady [299]

<h3>Answer:</h3>

[C] Velocity.

<h3>Explanation:</h3>

<u>As we know that</u>,

a = v - u/t

<u>where, a = acceleration, v = final velocity, u = initial velocity and t = time taken to travel</u>.

8 0

3 years ago

Read 2 more answers

A dog, that has a mass of 16 kgs, runs across a yard. What is the average force applied to the dog from the ground as it runs ac

VikaD [51]

Answer:

156.96 N

Explanation:

F=ma where m is the mass and a is acceleration

Substituting 16 Kg for m and 9.81 m/s2 for g then

F=16*9.81= 156.96 N

4 0

3 years ago

What is an equilibrant

professor190 [17]

The question seems to be what is an equilibrant force.

The answer is "an added force that produces equilibrium.

Here you have more insight:

<span>an object that has no net force acting on it? This object indeed is in equilibrium but the object is not the equilibran force.

the reaction force in an action-reaction pair of forces?

the reaction force is not an equilibrant force. The reaction force exists always but equilibrium is only possible if the net force is cero.

an added force that produces equilibrium? this is the right answer.</span>

5 0

3 years ago