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nasty-shy [4]
3 years ago
9

State one way to decrease the moment of a given force about a given axis of rotation.​

Physics
2 answers:
wlad13 [49]3 years ago
8 0

Answer:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

LiRa [457]3 years ago
5 0

Answer:

hope it helps...

Explanation:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

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anyanavicka [17]

Answer: 5m/L^2

Explanation:

Inertial I = mr^2 where r = distance from axis of rotation, while m is the mass of the object.

I = 2[m(1L/2)^2] + 2[m(3L/2)^2] = 2m×. 25/L^2+ 3m×2. 25/L^2= 0. 5m/l^2 +4. 5m/l^2

= 5m/l^2.

8 0
3 years ago
Two blocks are connected as shown in the diagram below. Assume that the ramp is frictionless. Draw the force diagram for the blo
cluponka [151]

Answer:

diagram: see image, x-component: 84.3 N, acceleration: 4.38 m/s^2

Explanation:

(see image for further explanation)

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3 years ago
How well a student reads can most likely be effected by
Burka [1]
Their cognitive skills and their ability to learn
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3 years ago
Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran
alina1380 [7]

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

3 0
3 years ago
A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.
Nastasia [14]
<h3>Solution for the above question : -</h3>

Ohm's law states that :

  • v = ir

the terms used are :

  • r = resistance
  • v = potential \:  \: difference
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let's solve for electric current :

  • 2 = i \times 40

  • i =  \dfrac{2}{40}

  • i = 0.05 \: A

  • i = 50 \: mA

\mathfrak{good\:  \: luck \:  \: for \:  \: your \:  \: assignment}

8 0
3 years ago
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