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3 years ago

Two vectors are shown. Which statement best compares the vectors?

Physics

2 answers:

DiKsa [7]3 years ago

8 0

Answer:

Explanation:

Because vector Y is longer than vector X

so when you take a magnitude( without minus ) each other

you see that Y>X

Leona [35]3 years ago

6 0

Answer:

Vectors Y has greater magnitude than vector X

Explanation:

The attached figure shows two vectors. A vector have both magnitude and direction.

The length of vector X is less than vector Y. They both are pointing in opposite directions.

The magnitude of a vector is given by :

a = xi +yj

$|a|=\sqrt{x^2+y^2}$

From the figure, it is clear that the vector Y has greater magnitude than vector X. Hence, the correct option is (B).

You might be interested in

Extrusive rocks forms beneath earth's surface true or false

Verizon [17]

The answer is false your welcome

8 0

4 years ago

10. What is the acceleration of a 1000 kg car subject to a 500 N net force?

Simora [160]

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

3 0

2 years ago

A truck is traveling at 2.0 m/s. It slows to a stop at a constant rate over 3.00 s. How far does the car travel during those 3.0

earnstyle [38]

Answer:

During those 3.00 seconds before stopping, the car travels a distance of 6 m.

Explanation:

The simple rule of three is a tool that is used to quickly solve problems, where three pieces of information must be known, and one of them operates as an unknown to be known.

Two magnitudes are directly proportional if one magnitude increases the other also does it, and if the magnitude decreases the other in the same way.

Being a, b and c known data and x the unknown, the value that we want to know, the rule of three when the magnitudes are directly proportional is applied as follows:

a ⇒ b

c ⇒ x

So: $x=\frac{c*b}{a}$

In this case, knowing that a truck travels at 2 m/s, the rule of three applies as follows: if in 1 second the truck travels 2 m, in 3 seconds how much distance does it travel?

$distance=\frac{3 s*2 m}{1 s}$

distance= 6 m

During those 3.00 seconds before stopping, the car travels a distance of 6 m.

8 0

3 years ago

An object starts from rest and uniformly accelerates at a rate of 1.25 m/s2 for 7.0 seconds.

stich3 [128]

Explanation:

Since its accelerating, the velocity vs time graph is linear

For displacement we need initial velocity (which is zero because it starts from rest) and final velocity (which is calculatee thro acceleration formula

A= (vf - vi)/t

a= vf-0/t

1.25=vf / 7

1.25*7=vf

8.75 = vf

Now for displacement plug all the values in

X = 1/2(vf-vi)/t formula

The displacement (x) is 30.625 m

For part 3, we know new displacement that is 22m , the final and initial velocities are the same so just plug in the values for same formula above

The answer is t = 5.02

Im pretty sure all the answers are correct

8 0

2 years ago

Please someone help, I’m very confused and it’s due soon, thanks

Anit [1.1K]

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

Additional comment

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0

3 years ago