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Liono4ka [1.6K]
3 years ago
6

In order to increase the speed in a gear system

Physics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

the driving gear must be larger than the driven gear

Explanation:

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Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k
IrinaVladis [17]

Answer:

magnitude = 3

unit vector = \frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}

Explanation:

Given vectors:

u = 2i + 2j - k

v = -i + k = -i + 0j + k

(a) u x v is the cross product of u and v, and is given by;

u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right]

u x v = i(2+0) - j(2 - 1) + k(0 - 2)

u x v = 2i - j - 2k

Now the magnitude of u x v is calculated as follows:

| u x v | = \sqrt{2^2 + (-1)^2 + (-2)^2}

| u x v | = \sqrt{4 + 1 + 4}

| u x v | = \sqrt{9}

| u x v | = 3

Therefore, the magnitude of u x v is 3

(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e

û = \frac{u X v}{|u X v|}        

u x v = 2i - j - 2k        [<em>calculated in (a) above</em>]

|u x v| = 3                   [<em>calculated in (a) above</em>]

∴ û = \frac{2i - j - 2k}{3}

∴ û = \frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}

4 0
3 years ago
Study the images above. Choose all of the statements below that are true regarding wave energy.
Dafna1 [17]

Answer:

1) Letter J identifies the wavelength of the wave.

4) If letter G was longer, The wave would have more energy.

6) The wavelength of a microwave is larger than that of an x-ray.

Explanation: Just did it on USA Test Prep

3 0
3 years ago
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A 1.50-kg red cart is moving rightward with a speed of 60.0 cm/s when it collides with a 0.50-kg blue cart that is moving leftwa
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Answer:

byd5tre r rte566ed6rtsdfcffdrdrfxvc

Explanation:

7 0
4 years ago
Object A of mass M is released from height H, whereas object B of mass 0.5M is released from height 2H. What is the ratio of the
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My solution. ......................

5 0
3 years ago
A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds af
Vilka [71]

Answer:

V = (-32t +32) ft/s

The velocity of the bowling ball after t seconds is the time derivative of the position function s(t). To obtain the velocity from s(t) we simply differentiate s(t) with respect to t. That is

V = ds(t)/dt = -16×2t + 32×1 = -32t + 32.

When differentiating, you multiply the coefficient of each term in the equation with the power of the variable and then reduce the power by 1 just like above.

Explanation:

8 0
3 years ago
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