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Katen [24]
3 years ago
15

Why might you want to use a single fixed pulley?

Physics
1 answer:
hjlf3 years ago
4 0

A single fixed pulley can be used to raise or lower lightweight objects.

Option b

<u>Explanation:</u>

A pulley is a simple machine tool which is used to make lifting or lowering tasks easy.  A single fixed pulley is a system involving only one pulley fixed on a constant rigid support with a rope wrapped around the wheel. Such a system can be used only to change the direction of  applied force in raising or lowering small, lightweight objects which need minimal work force.

A single fixed pulley system helps only in redirecting the applied force direction by using a rope and wheel assembly. The work done in such a case remains the same and hence it is not preferred to use it in lifting heavy objects. Neither is the required force reduced in case of a single fixed pulley system. A movable pulley helps in achieving (A) and (C).

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We know, density = Mass / Volume
It represents density is directly proportional to mass and indirectly proportional to volume. So, at a constant volume, object with larger density will appear more heavier than that of object with smaller density.

In short, Heavier object will have the higher density than the other.

Hope this helps!
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You are doing a measurement to determine the conductance of an artificial membrane. The membrane is selectively permeable to Na
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Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f
musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

E_1_x = 0V/m\\E_1_y=100V/m

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m

∑x: E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m

∑y: E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees

or 27.23° to the left of E1.

8 0
3 years ago
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