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lana66690 [7]
3 years ago
6

Vector A⃗ points in the positive y direction and has a magnitude of 12 m. Vector B⃗ has a magnitude of 33 m and points in the ne

gative x direction.
Find the direction of A⃗ + B⃗ .
Find the magnitude of A⃗ + B⃗
Physics
1 answer:
gavmur [86]3 years ago
6 0

vector A has magnitude 12 m and direction +y

so we can say

\vec A = 12 \hat j

vector B has magnitude 33 m and direction - x

\vec B = -33 \hat i

Now the resultant of vector A and B is given as

\vec A + \vec B = 12 \hat j - 33 \hat i

now for direction of the two vectors resultant will be given as

\theta = tan^{-1}\frac{12}{-33}

\theta = 160 degree

so it is inclined at 160 degree counterclockwise from + x axis

magnitude of A and B will be

R = \sqrt{A^2 + B^2}

R = \sqrt{12^2 + 33^2} = 35.11 m

so magnitude will be 35.11 m

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The total displacement of the toy car at the given positions is 0.

The given parameters;

  • <em>First displacement of the car, = 5 cm left</em>
  • <em>Second displacement of the car, = 8 cm right</em>
  • <em>Third displacement of the car, = 3 cm to the left</em>

The total displacement of the car is calculated as follows;

  • <em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
  • <em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>

\Delta x = - \ 5\ cm  \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0

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Learn more about displacement here: brainly.com/question/18158577

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A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

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Answer:

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Gneiss has resistance to pressure and mechanical impacts

Explanation:

Granite is an igneous rock. It is mostly used in building works and construction because they are very durable. They are hard and tough and they have no internal structures.

Gneiss is used for flooring, ornamental stone, tombstones because of the fact that it shows resistances to pressure and also mechanical impacts.

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In nature, granite is formed from the cooling down of hot molten magma and it's solidification before it reaches the surface of the earth.

In nature, gneiss is as a result of igneous rock or sedimentary rocks metamorphosing. Gneiss and granite are kind of similar. When subjected to great heat, granite becomes gneiss

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