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garik1379 [7]
3 years ago
15

A long straight rod experiences several forces, each acting at a different location on the rod. All forces are perpendicular to

the rod. The rod might be in translational equilibrium, rotational equilibrium, both, or neither.
(1) If a calculation reveals that the net torque about the left end is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net force on the rod is also zero.
(C) might not be in rotational equilibrium even if the net force on the rod is also zero.
(D) might be in rotational equilibrium even if the net force is not zero.
(2) If a calculation reveals that the net force on the rod is zero, then one can conclude that the rod
(A) is definitely in rotational equilibrium.
(B) is in rotational equilibrium only if the net torque about every axis through anyone point is found to be zero.
(C) might be in rotational equilibrium if the net torque about every axis through anyone point is found to be zero.
(D) might be in rotational equilibrium even if the net
torque about any axis through anyone point is not zero.
Physics
1 answer:
enot [183]3 years ago
6 0

Answer:

1. C

2.C

Explanation:

1. The rod is perpendicular to every axis and forces are acting on every location. If the torque on the left side is zero, this indicates that forces with respect to their distance on the left side is zero and doesn't account for the net force at a point.

2. If the net torque about every point on every axis is zero, the rod will be rotational because each axis will yield a magnitude of zero which obeys the principle of rotation at a point.

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The phases of the moon depend on how much of the lighted side of the moon can be seen from earth.is this true or false
Sindrei [870]
The answer to this question is going to be False 

6 0
3 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
A circuit consists of a battery connected to three resistors (65 ω, 25ω, and 170ω) in parallel. the total current through the re
White raven [17]
A. To find the total emf of the battery, just remember that in a parallel circuit, the voltage is the same throughout the circuit. So you can get the total voltage of the circuit by using Ohm's Law. 

I= \frac{V}{R}

Where:
I = current (A)
V = Voltage (V) (emf)
R = Resitance (Ω)

Now you can derive the formula of Voltage by transposing the Resistance to the other side of the equation to isolate Voltage. The formula you will now use will be:
V = IR

However, you cannot solve this yet because the resistance you need is the total resistance in the circuit. To do this, you need to get the total resistance in this parallel circuit and the formula would be:

\frac{1}{R_{T}} =  \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}


\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059
\frac{1}{R_{T}} =0.0613

Get the reciprocal of both sides and divide:

R_{T} =  \frac{1}{0.0613} =16.32

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
V = IR
V = (1.8)(16.32)
V = 29.376V

The emf of the battery is 29.376V


B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
I= \frac{V}{R}
I= \frac{29.376}{65}
I= 0.45A

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
I= \frac{V}{R}
I= \frac{29.376}{25}
I= 1.18A
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
I= \frac{V}{R}
I= \frac{29.376}{170}
I= 0.17A

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.
4 0
2 years ago
Select the statement that best describes gravity.
pav-90 [236]

Explanation:

Gravity is the force of attraction between two objects. It depends upon the mass of the objects and the distance between the objects. Mathematically, the force of gravity is given by :

F=G\dfrac{m_1m_2}{d^2}

Where

G is the universal gravitational constant

m_1\ and\ m_2 are masses

d is the distance between two masses

So, statement (2) describes gravity "Gravity is the force of attraction between two objects; it is dependent upon the mass of the objects and the distance between the objects".

6 0
3 years ago
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mamaluj [8]

Answer:

c

Explanation:

Science influences society through its knowledge and world view. ... Technology influences society through its products and processes

6 0
2 years ago
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