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3 years ago

What does this mathematical expression mean? D 1/T

2 answers:

7 0

The question is incomplete. If D=1/T then the answer is D, but if instead of =, there is a proportional sign, then the answer is C.

notsponge [240]3 years ago

6 0

Answer: density Is inversely related to temperature

Explanation:

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750 million years ago, there was enough Uranium-235 present in Oklo, Gabon to have a natural fission reactor occur and generate

Komok [63]

<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>

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4 years ago

Read 2 more answers

Describe how air pressure is exerted on objects.

Delicious77 [7]

<span>The same amount of pressure is exerted on all sides of the object equally. If you had 13 psi, then that means on the top of the object, 13 psi would be pushing in on the object. From the bottom, 13 psi would be pushing up on the object. Same goes for all the sides as well. The pressure pushes 'in' on the object.</span>

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3 years ago

A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.

aleksley [76]

Answer:

w = 25.05 rad / s , α = 0.7807 rad / s² , θ = 1972.75

Explanation:

This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off

θ = w₀ t - ½ α t²

α = (w₀t - θ) 2/t²

let's reduce the magnitudes to the SI system

w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s

θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad

let's calculate the angular acceleration

α = (55.5 39 - 1570.8) 2/39²

α = 0.7807 rad / s²

having the acceleration we can calculate the final speed

w = w₀ - ∝ t

w = 55.5 - 0.7807 39

w = 25.05 rad / s

the time to stop w = 0

0 = wo - alpha t

t = wo / alpha

t = 55.5 / 0.7807

t = 71.09 s

the angle traveled

w² = w₀⁹ - 2 α θ

w = 0

θ = w₀² / 2α

let's calculate

θ = 55.5 2 / (2 0.7807)

θ = 1972.75

5 0

3 years ago

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

3 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$