How does an atom of bromine-79 become a bromide ion with a -1 charge?

Water flows into a swimming pool at the rate of 5.85 gal/min. If the pool dimensions are 21.2 ft wide, 46.1 ft long and 19.4 ft

Lana71 [14]

First, calculate volume:

volume = 21.2 ft * 46.1 ft * 19.4 ft

volume = 18,960 ft^3

Convert to gallon:

volume = 18,960 ft^3 * (12 in / 1 ft)^3 * (1 gallon / 231 in^3)

volume = 141,830.71 gallon

Therefore the time is:

time = 141,830.71 gallon / (5.85 gal/min)

time = 24,244.57 min

6 0

3 years ago

The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.5

MrRissso [65]

Question is not complete and the missing part is;

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

Answer:

0.828 m/s

Explanation:

Resolving vertically, we have;

Fn and Fg act vertically. Thus,

Fn - Fg = 0 - - - - eq(1)

Resolving horizontally, we have;

Ff = ma - - - - eq(2)

Now, Fn and Fg are both mg and both will cancel out in eq 1.

Leaving us with eq 2.

So, Ff = ma

Now, Frictional force: Ff = μmg where μ is coefficient of friction.

Also, a = v²/r

Where v is linear speed or velocity

Thus,

μmg = mv²/r

m will cancel out,

Thus, μg = v²/r

Making v the subject;

rμg = v²

v = √rμg

Plugging in the relevant values,

v = √0.14 x 0.5 x 9.8

v = √0.686

v = 0.828 m/s

3 0

3 years ago

Derived and define the units

Vadim26 [7]

Dont click on the link its a scam btw

8 0

3 years ago

A beetle with a mass of 15.0 g is initially at rest on the outer edge of a horizontal turntable that is also initially at rest.

vichka [17]

Answer:

- 8.33 x 10⁻³ rad /s ( anticlockwise)

Explanation:

The rotational movement of beetle and turntable is caused by torque generated by internal forces , we can apply conservation of angular momentum.

That is ,

I₁ ω₁ = I₂ω₂ , ω₁ and ω₂ are angular velocity of beetle and turntable respectively.

ω₁ + ω₂ = .05 radian /s ( given )

Momentum of inertia of beetle I₁ = mass x (distance from axis)²

= 15 x 10⁻³ x R² ( R is radius of the turntable )

Momentum of inertia of turntable I₂ =1/2 mass x (distance from axis)²

= 75/2 x 10⁻³ x R² ( R is radius of the turntable )

I₁ ω₁ = I₂ω₂ ,

15 x 10⁻³ x R² x ( .05 - ω₂ ) = 75/2 x 10⁻³ x R² ω₂

15 x ( .05 - ω₂ ) = 75/2 x ω₂

.75 - 15ω₂ = 37.5ω₂

.75 = 52.5 ω₂

ω₂ = - 14.3 x 10⁻³ rad /s ( anticlockwise)

7 0

3 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$