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Alja [10]
3 years ago
9

Three vectors →a, →b, and →c each have a magnitude of 50 m and lie in an xy plane. Their directions relative to the positive dir

ection of the x axis are 30°, 195°, and 315°, respectively. What are (a) the magnitude and (b) the angle of the vector →a+→b+→c and (c) the magnitude and (d) the angle of →a−→b+→c? What are the (e) magnitude and (f) angle of a fourth vector →d such that (→a+→b)−(→c+→d)=0 ?
Physics
1 answer:
Damm [24]3 years ago
5 0

Answer:

a) 38.27      b) 322.5°

c) 126.99    d) 1.17°

e) 62.27     e) 139.6°

Explanation:

First of all we have to convert the coordinates into rectangular coordinates, so:

a=( 43.3 , 25)

b=( -48.3 , -12.94)

c=( 35.36 , -35.36)

Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):

1.)  a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°

2.)  a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°

3.)  (a+b) - (c+d)=0   Solving for d:

     d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°

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All of the following processes are the ones that lead directly to the formation of sedimentary rock EXCEPT-
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A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist
Paha777 [63]

Answer:

Part a)

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

\frac{1}{2}mv^2 = mgh'

now we know that

I = \frac{2}{5}mr^2

\omega = \frac{v}{r}

so we have

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh

\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh

\frac{7}{10}mv^2 = mgh

\frac{1}{2}mv^2 = \frac{10}{14}mgh

so the height on the smooth side is given as

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

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4 years ago
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