All categories

3 years ago

Does the law of attraction actually have a basis in physics?

Physics

2 answers:

Tanya [424]3 years ago

7 0

That is correct

There is not answer choices but I think its true

Are you good with compare and contrasting?

If so, look at my recent question please

allochka39001 [22]3 years ago

7 0

This statement is true

You might be interested in

3 CO2<br> How many atoms of Oxygen (O) are there?

svp [43]

Pretty sure there are 2 since you’d normally write it with the 2 below the O symbolizing the number of atoms

8 0

3 years ago

Read 2 more answers

Chris threw a basketball a distance of 27.5 m to score and win his

salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

$R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}$

8 0

3 years ago

Describe three common issues that affect societies across places and times

Vesna [10]

Poverty, Natural Disaster, and Disease are 3 answers.

6 0

3 years ago

Jake appears to have several personalities. One is his dead cousin Bob, who

bazaltina [42]

The answer is C. Dissociative identity disorder

-

This should be easy to spot

8 0

3 years ago

Read 2 more answers

f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$