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3 years ago

Where does the energy come from?

Physics

1 answer:

Semenov [28]3 years ago

5 0

Answer: Scientist don't know where energy comes from and how it was created but God created energy.

Explanation:

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The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug

9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0

4 years ago

Read 2 more answers

When it reacts with another element, an oxygen atom gains two electrons.

luda_lava [24]

Answer:

Explanation:

An ion with a negative charge

3 0

3 years ago

Read 2 more answers

A 3,000 kg truck moving at +10 m/s hits a 1,000 kg parked car which moves off at +15 m/s What is the velocity of

Rina8888 [55]

Answer:

v₃ = 5 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.

P = m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision

Pbeforecollision = Paftercollision

(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the truck = 3000 [kg]

v₁ = velocity of the truck = 10 [m/s]

m₂ = mass of the car = 1000 [kg]

v₂ = velocity of the car before the collision = 0 (the car is parked)

v₃ = velocity of the truck after the collision [m/s]

v₄ = velocity of the car after the collision = 15 [m/s]

(3000*10) + (1000*0) = (3000*v₃) + (1000*15)

30000 = 3000*v₃ + 15000

3000*v₃ = 30000 - 15000

3000*v₃ = 15000

v₃ = 5 [m/s]

7 0

3 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

3 years ago

Read 2 more answers

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

4 years ago