All categories

2 years ago

1. Two vectors have magnitudes 6 & 8 units, respectively. Find the maximum &

Physics

1 answer:

defon2 years ago

5 0

Answer:

<em>The maximum magnitude of their resultant is 14 units, and the minimum is 2 units</em>

Explanation:

<u>Resultant Vectors</u>

When two or more vectors are added or subtracted, the resultant vector can be found by considering their magnitudes and directions.

Two vectors applied to the same point can produce a result that can vary from being completely collaborative or completely opposite.

If two vectors act in a collaborative form, their magnitudes are added and the result has the maximum possible magnitude. If they act in opposite directions, the result has the minimum possible magnitude.

Thus, being 6 and 8 units the magnitudes of the vectors, the maximum magnitude of their resultant is 14 units, and the minimum is 2 units.

You might be interested in

While drag racing out of the parking lot I timed myself at a speed of

VladimirAG [237]

Answer:

maybe ill be tracer

Explanation:

3 0

3 years ago

A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t

Gelneren [198K]

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04 x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

$V=\dfrac{4}{3}\pi r^3$

$V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3$

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb + N = m g

m=Mass of the ball = Density x volume

m = γ V , γ =Density of the Ball

ρ V g + N = γ V g ( take g= 10 m/s²)

$\gamma =\dfrac{N+\rho V g}{V g}$

$\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times 10}{219.13\times 10}$

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

5 0

2 years ago

Calculate the acceleration of a galloping horse going from 2 m/s to 12 m/s in 2 seconds.

olchik [2.2K]

A)
5m/s^2

(12m/s-2m/s)
__________ = 5m/s^2
2s

8 0

3 years ago

Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

3 years ago

VERY EASY QUESTION!!!!!! 20 POINTS!!!!!!!!

slamgirl [31]

C. at a science fair

4 0

2 years ago

Read 2 more answers