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2 years ago

13) Can two nonzero perpendicular vectors be added together so their sum is zero? Explain.

Physics

2 answers:

statuscvo [17]2 years ago

4 0

Answer:

sum of two perpendicular nonzero vectors can not be ZERO

Explanation:

As we know by the sum or addition of two vectors is given by

$R = \sqrt{a^2 + b^2 + 2abcos\theta}$

here we know that

a and b are the magnitude of two vectors

$\theta$ = angle between two vectors

so as we know that here two vectors are perpendicular to each other

so we will have

$R = \sqrt{a^2 + b^2}$

now since the is sum of two positive non zero numbers so it can not be zero

so sum of two perpendicular nonzero vectors can not be ZERO

Neporo4naja [7]2 years ago

3 0

In order for two vectors to add to zero, they must have the same magnitude and point in opposite directions.

Two perpendicular vectors, by definition, make a right angle with each other whereas two vectors pointing in opposite directions form a straight line.

Because of this, two perpendicular vectors with nonzero magnitudes will never add to zero.

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A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

1 year ago

Read 2 more answers

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing

melomori [17]

Answer: Anthony will be warmer after the game.

Explanation :

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing a white shirt. Anthony will be warmer after the game. The black color is a good absorber of radiation and a bad reflector.

The black color absorbs heat until a thermal equilibrium is attained. So, it is advisable to wear cotton clothes in summers not dark colored clothes.

7 0

2 years ago

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A golfer hits a golf ball at an angle of 25.0Â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is

kvasek [131]

<u>Answer:</u>

Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Vertical motion (Maximum height reached, H) :

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.