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Dmitriy789 [7]
3 years ago
15

Find the vector parallel to the resultant of the vector A=i +4j-2k and B=3i-5j+k​

Physics
2 answers:
Zinaida [17]3 years ago
8 0

Answer:

2008

Explanation:

2000+3+5======2008

valkas [14]3 years ago
6 0

Answer:

8\hat i-2\hat j-2\hat k

Explanation:

<u>Vectors in 3D</u>

Given a vector

\vec P = P_x\hat i+P_y\hat j+P_z\hat k

A vector \vec Q parallel to \vec P is:

\vec Q = k.\vec P

Where k is any constant different from zero.

We are given the vectors:

\vec A = \hat i+4\hat j-2\hat k

\vec B = 3\hat i-5\hat j+\hat k

It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:

\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k

Adding each component separately:

\vec A +\vec B = 4\hat i-\hat j-\hat k

To find a vector parallel to the sum, we select k=2:

2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k

Thus one vector parallel to the resultant of both vectors is:

\mathbf{8\hat i-2\hat j-2\hat k}

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The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff
stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

where;

\mathbf { N_A  \ and  \ N_B} are the normal contact force at center point A and B respectively .

N_A+N_B = 45*9.8

N_A+N_B = 441    ------- equation (1)

Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

\mathbf{ F_A \ and \ F_B } are the static friction at center point A and B respectively.

which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

P = 0.30 ( 441)    

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

P*h = mg*0.24

h =\frac{45*9.8*0.24}{132.3}

h = 0.8 m

Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

6 0
3 years ago
Help please (20 pts) I think it's C
gregori [183]

Answer:

I'm not sure it is c I'm sure it is d

6 0
3 years ago
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solniwko [45]

Answer:

Explanation:

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5 0
3 years ago
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What happens to the coefficient of friction when the weight is increased? Why is this?
Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0
3 years ago
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Andrei [34K]

Answer:

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Explanation:

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