Which of the following is the brightness of a star as we see if from Earth?

A long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and

natima [27]

The upward force exerted on the board by the support is 530.8 N.

<h3>Upward force exerted on the board by the support</h3>

The sum of the upward forces is equal to sum of downward forces;

total downward forces = 52.8 N + 206 N + 272 N = 530.8 N

downward force = upward force = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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8 0

1 year ago

At which location could you place the south pole of a bar magnet so that it would be pulled toward the magnet shown?

stepan [7]

Answer:

The north pole.

Explanation:

3 0

3 years ago

An astronaut on a strange new planet having no atmosphere finds that she can jump up to a maximum height of 27 m when her initia

vekshin1

The magnitude is 4.5

7 0

3 years ago

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This is Yolanda's desk. What could Yolanda do to increase the amount of force needed to change the motion of the desk?

Stella [2.4K]

Yolanda might put more items to the desk to make it heavier, requiring more force.

We need to learn more about the force acting on an object in order to locate the solution.

<h3>How can the force that is required to modify the motion be increased?</h3>

We are aware that the word for force is,

F=ma

where m denotes the object's mass and an its acceleration

There are two ways to increase the force required to alter the motion of the table.

One is to increase the mass, and the other is to accelerate it more quickly.

Otherwise, there will be a lot of friction between the surfaces, making it difficult to move without exerting a lot of force.

We can infer from this that Yolanda could add items to the desk to increase its mass, necessitating the use of additional force.

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8 0

1 year ago

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 9

Ivanshal [37]

Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

$P=\frac{90I}{I^2+I+9}$

now differentiating the above equation with respect to Intensity 'I' we get

$\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}$

Now for the maxima $\frac{dP}{dI}=0$

thus,

$0=\frac{-90I^2+810)}{(I^2+I+9)^2}$

or

$-90I^2+810=0$

or

$I^2=\frac{810}{90}$

or

$I^2=9$

or

I = 3

thus, <u>for the value of intensity I = 3, the P is maximum</u>

at I = 3

$P=\frac{90\times3}{3^2+3+9}$

or

$P=\frac{270}{21}$

or

$P=12.85$

5 0

3 years ago