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2 years ago

14

Psuedopod in a sentence

1 answer:

mixas84 [53]2 years ago

5 0

Does it have to be that exact word. cause it is just another term for psuedopodium

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A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

MissTica

Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

5 0

3 years ago

Can you help me with this paper please I will give you 20 points!

sweet [91]

1) they are attracting because if you look at the arrows they’re all pointing the same way.

2) if the magnet was turned around they would do the opposite and not attract ( this is called repulsion)

3) magnetic pole

4)magnet

5) magnetic force

6) magnetism

Hope this helps

6 0

2 years ago

Read 2 more answers

MA of the first class lever may be equal to, greater than 1.Why

Anika [276]

Answer:

hjhjhjhjhjhj

Explanation:

jmjk

4 0

3 years ago

Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A

Finger [1]

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

$r=(100\cos2\theta)\ m$

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

$v=r\omega$

$\omega=\dfrac{v}{r}$

Where, v = velocity

r = radius

Put the value into the formula

$\omega=\dfrac{29.0576}{100\cos30}$

$\omega=0.335\ rad/s$

Hence, The rate at which radar must rotate is 0.335 rad/s.

3 0

3 years ago

An airplane is pushed 22 degrees eastward off its northward course by a jet stream traveling at a speed of 136.73 km/hr. The new

uranmaximum [27]

As the speed of airplane is change due to jet stream

So the net speed is given as

$v_{net} = v_{plane} + v_{stream}$

now we can rearrange it as

$v_{plane} = v_{net} - v_{stream}$

now by the formula of vector difference we have

$v_{plane} = \sqrt{v_{net}^2 + v_{stream}^2 - 2v_{net}v_{stream}cos\theta}$

now plug in all values

$v_{plane} = \sqrt{365^2 + 136.73^2 - 2* 365* 136.73*cos22}[tex]v_{plane} = 243.7 km/hr$

so above is the speed of the plane

3 0

2 years ago