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valentinak56 [21]
3 years ago
15

Now, consider the resultant electric field e⃗ net at p. with reference to the coordinate system shown in the previous part, whic

h component of e⃗ net, if any, is zero in this problem?
Physics
1 answer:
Rzqust [24]3 years ago
5 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices:
only the x component
only the y component
both the x and y components
neither the x nor the y component

The answer is neither the x nor the y component

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Take a look at the weather map. The front seen there causes short periods of storms and heavy rains. What type of front is this?
erastova [34]
I am pretty sure it is a cold front.
5 0
3 years ago
Read 2 more answers
An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len
worty [1.4K]

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

7 0
3 years ago
A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr
Papessa [141]

Answer:

The  moment of inertia about an axis through the center and perpendicular to the plane of the square is

    I_s =  \frac{Ma^2}{3}

Explanation:

From the question we are told that

   The length of one side of the square is  a

   The total mass of the square is  M

Generally the mass of one size of the square is mathematically evaluated as

    m_1 = \frac{M}{4}

Generally the moment of inertia of one side of the square is mathematically represented as

        I_g =  \frac{1}{12}  *  m_1 * a^2

Generally given that m_1 = m_2 = m_3 = m_4 = m it means that this moment inertia evaluated above apply to every side of the square  

Now substituting for  m_1

  So

       I _g=  \frac{1}{12}  *  \frac{M}{4} * a^2

Now according to  parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_a =  I_g + m [\frac{q}{2} ]^2

=>    I_a =  I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2

substituting for I_g

=>    I_a =  \frac{1}{12}  *  \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2

=>    I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}

=>    I_a = \frac{Ma^2}{12}

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

      I_s = 4 * I_a

=>   I_s = 4 * \frac{Ma^2}{12}

=>   I_s =  \frac{Ma^2}{3}

8 0
3 years ago
A force is applied to a 30-kg object, resulting in an acceleration of 15 m/s. What is the force being applied to the object?
telo118 [61]
The sum of all forces applied to the object = (mass) x (acceleration)

The sum of all forces applied to the object = (30) x (15)

= <u>450 newtons</u>, in the direction of the acceleration.
 
5 0
3 years ago
Read 2 more answers
The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-
Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a  \\ \\ a = \dfrac{347.906}{50} \\ \\ a  = 6.96 \ m/s^2

Bending moment M:

From the second part of the diagram:

\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435  \\ \\  M = 238.435 + 220.725 \\ \\  \mathbf{M = 459.16 \ N.m}

6 0
3 years ago
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