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3 years ago

Show that A = î- 5 j are parallel to each other.B= 2î-10 j are parallel to each other.

Physics

1 answer:

Thepotemich [5.8K]3 years ago

3 0

We are given:

2 Vectors A and B

A = î - 5j

B = 2î - 10j

Proving that the given vectors are Parallel:

We can rewrite the Vector B as: B = 2(î - 5j)

Since A = î - 5j

Vector B can be written as:

B = 2(A vector)

this means that the magnitude of Vector B is twice of vector A but their direction is the same

Since the 2 vectors are moving in the same direction, they are parallel

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A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration

Dimas [21]

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

$v_{o,x}$ - Initial x-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$ , $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

$v_{o,y}$ - Initial y-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o,y} = -3.6\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}$

$y(t) = -2.16\,m$

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ( $v_{x}$ ), in meters per second, is calculated by this kinematic formula:

$v_{x}(t) = v_{o,x} + a_{x}\cdot t$ (3)

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-velocity of the skateboarder is:

$v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)$

$v_{x}(t) = 1.08\,\frac{m}{s}$

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

$v_{y} = -3.6\,\frac{m}{s}$

The y-velocity of the skateboard is -3.6 meters per second.

3 0

3 years ago

Suppose the B string on a guitar is 24" long and vibrates with a frequency of about 247 Hz. You place your finger on the 5th fre

pashok25 [27]

To solve this problem it is necessary to apply the related concepts to string vibration. This concept shows the fundamental frequency of a string due to speed and length, that is,

$f = \frac{v}{2L}$

Where

v = Velocity

L = Length

Directly if the speed is maintained the frequency is inversely proportional to the Length:

$f \propto \frac{1}{L}$

Therefore the relationship between two frequencies can be described as

$\frac{f_2}{f_1}=\frac{l_1}{l_2}$

$f_2 = \frac{l_1}{l_2}(f_1)$

Our values are given as,

$l_1 = 24"\\f_1 = 247Hz\\l_2 = 18"$

Therefore the second frequency is

$f_2 = \frac{l_1}{l_2}(f_1)\\f_2 = \frac{24}{18}(247)\\f_2 = 329.33Hz$

The frequency allocation of 329Hz is note E.

8 0

3 years ago

A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

12345 [234]

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

v = √Rg

v = √(1.0(9.8))

v = 3.1304951...

v = 3.1 m/s

6 0

3 years ago

Which of the following accurately describes the way in which a muscle moves?

Vaselesa [24]

<h3>Answer;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3>Explanation;</h3>

The events of muscle fiber shortening occurs with in the sacromeres in the fibers.
Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.
Filament movement starts at the region or zone where thin and thick filaments overlap.
Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.

7 0

3 years ago

hot-air balloon is ascending at the rate of 14 m/s and is 84 m above the ground when a package is dropped over the side. (a) How

timurjin [86]

Answer:

a) t = 4.14 s

b) Speed with which it hits the ground = 40.58 m/s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = 84 m,

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) y = ut + gt²/2

84 = 0 + 9.8t²/2

4.9t² = 84

t² = 84/4.9

t = 4.14 s

b) v = u + gt

v = 0 + (9.8 × 4.14)

v = 40.58 m/s

4 0

4 years ago

Read 2 more answers

Show that A = î- 5 j are parallel to each other.B= 2î-10 j are parallel to each other.​

Show that A = î- 5 j are parallel to each other.B= 2î-10 j are parallel to each other.