Ask question

Ask question

All categories

3 years ago

5

The Pacific Plate is moving 29 mm/year toward the north and 20 mm/year toward the west relative to the North American Plate. Sho

w how to use geometry and the Pythagorean Theorem to calculate the total relative motion toward the northwest.

1 answer:

adell [148]3 years ago

4 0

Answer:

7 mm per year

Explanation:

It is given that :

The Pacific plate is moving towards north at = 29 mm per year

The Pacific plate is moving towards west at = 20 mm per year

We have to calculate the total relative motion towards the northwest.

So we have to find the resultant of the two motions.

Since the two movements are perpendicular, therefore the angle between the two motions is 90 degree.

Therefore, finding their resultant,

$R^2=(29)^2+(20)^2$

$R^2=(49)^2$

R = 7

Therefore, total relative motion towards the northwest is 7 mm per year.

You might be interested in

A block of mass 0.500 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x (F

Marrrta [24]

Answer:

Part a)

$x = 0.4 m$

Part b)

$v_i = 11.7 m/s$

Part c)

Speed is more than the required speed so it will reach the top

Explanation:

Part a)

As we know that there is no frictional force while block is moving on horizontal plane

so we can use energy conservation on the block

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

$\frac{1}{2}0.500(12^2) = \frac{1}{2}(450)x^2$

$x = 0.4 m$

Part b)

If the track has average frictional force of 7 N then work done by friction while block slides up is given as

$W_f = -7( \pi R)$

$W_f = -7(\pi \times 1.00)$

$W_f = -22 J$

work done against gravity is given as

$W_g = - mg(2R)$

$W_g = -(0.500)(9.8)(2\times 1)$

$W_g = -9.8 J$

Now by work energy equation we have

$\frac{1}{2}mv_i^2 + W_f + W_g = \frac{1}{2}mv_f^2$

$\frac{1}{2}0.5(12^2) - 9.8 - 22 = \frac{1}{2}(0.5)v_f^2$

$v_f = 4.1 m/s$

Part c)

now minimum speed required at the top is such that the normal force must be zero

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = 3.13 m/s$

so here we got speed more than the required speed so it will reach the top

5 0

3 years ago

1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground

Svetradugi [14.3K]

You have three known variables:

Acceleration - $9.8m/s^2$
Time - $1.85s$
Initial Velocity - $0m/s$

For the first part of your question:

$v = u + at

v = (0)+(9.8)(1.1)

v = 10.78 m/s v=11 m/s (2 significant figures)$

For the second part of your question:

$v=u+at

v=(0)+(9.8)(1.85)
v=18.13 m/s

$

This still needs to be converted to m/h:

$18.13m/s = 18.13\times 3600 metres/h=65628 metres/hour

65628 metres/hour = 65628\div1600 mi/h = 40.7925 mi/h

= 41 mi/hr (2 significant figures)
$

5 0

4 years ago

Read 2 more answers

Describe how charge is transferred from the ruler to the metal rod.

spin [16.1K]

When the ruler is broughı near the inetal knob, it repels electrons in the metal. Electrons move away froni the ruler and down the metal rod. The knob now has a positive charge. The thin pieces of metal foil at the bottom of the metal rod now have a negative charge.

3 0

3 years ago

Read 2 more answers

Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack s

Maru [420]

Answer:

a) speed when Jack sees the pot : 12.92 meters per second

b) height difference 163.115 meters

Explanation:

First to calculate te initial speed we use the acceleration formula:

a= v1-v0/t

Acceleration being gravity's acceleration (9.8 m/s^2)

v1 being the speed when Jill sees the pot

v0 when Jack sees it

and t the time between

Solving for v0 it would be

v1 - a*t = v0

replacing

$58 m/s - 9.8 m/s^2 *4.6 s = v0 = 12.92 m/s$

For the second question we use the position formula setting y0 and t0 as the position and time when jack sees the pot. (and setting the positive axis downward I.E. one meter below jack would be 1m not -1m)

The formula is

$y0 + v0*t + 1/2 g *t^2 = yt$

replacing

$0m + 12.92m/s* 4.6 s + 1/2 * 9.8 m/s^2 * (4.6 s)^2 = 163.115 m$

5 0

3 years ago

a 3520 kg truck moving north at 18.5 m/s makes an INELASTIC collision with an 1480 kg car moving east after colliding they have

anyanavicka [17]

Answer:

Explanation:

An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is

p = mv and

p = (3520 + 1480)(13.6) so

p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is

$p_f=\sqrt{(p_{truck})^2+(p_{car})^2}$ which, in words, is

the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:

$68000=\sqrt{(65100)^2+(1480v)^2}$ and

$(68000)^2=(65100)^2+(1480v)^2$ and

$4624000000=4238010000+2190400v^2$ and

$385990000=2190400v^2$ and

$176.2189554=v^2$ so

v = 13.3 m/s at 72.6°

6 0

3 years ago