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Harlamova29_29 [7]

2 years ago

5

2. Willingness to take turns is one way we can express our attitudes through

1 answer:

leva [86]2 years ago

3 0

B because willingness shows confidence in your answer

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Draw the following vector quantity Using the coordinate system.

DiKsa [7]

The given vectors quantities can be described by their properties of both

magnitude and direction.

a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.

b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.

c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

$\vec{F}$ = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; $\vec{v}$ = 120 × cos(25)·i + 120×sin(25)·j, which gives;

$\vec{v}$ = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

$\vec{v}$ ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is $\vec{d}$ = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

$\vec{d}$ = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

brainly.com/question/10409036

4 0

2 years ago

A projectile is fired with a velocity of 45 m/s at an angle of 32°. What is the

Dennis_Churaev [7]

Answer:

C

Explanation:

To calculate adjacent of triangle use

$\cos(32) = \frac{a}{45}$

where 45 is the hypotenuse, and a is the adjacent side (horizontal component)

$45 \cos(32) = 38.16$

then round to 38.2

8 0

3 years ago

please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ba

viva [34]

Answer:

the acceleration during the collision is: - 5 $\frac{m}{s^2}$

Explanation:

Using the formula:

$a=\frac{\Delta\,v}{\Delta\,t}$

we get:

$a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}$

4 0

3 years ago

A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

<u>Now the horizontal component of velocity:</u>

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

<u>The vertical component of the velocity:</u>

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

<u>Time taken by the projectile to travel the distance of 30 m:</u>

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

<u>Vertical position of the projectile at this time:</u>

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

<u>We observe that </u> $t>t'$ <u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point

tino4ka555 [31]

No matter what direction you throw it, or with what speed, its acceleration is immediately 9.8 m/s^2 downward as soon as you release it from your hand, and it doesn't change until the ball hits something.

4 0

3 years ago