In the rectangle to the left, if

A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because

amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0

3 years ago

The ground state energy of an oscillating electron is 1.23 eV. How much energy must be added to the electron to move it to the t

Vikki [24]

Answer:

The energy that must be added to the electron to move it to the third excited state is -1.153 eV

The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV

Explanation:

Given;

Energy of electron in ground state (n = 1 ) = 1.23 eV

E₁ = 1.23 eV

Eₙ = E₁ /n²

where;

E₁ is the energy of the electron in ground state

n is the energy level,

For third excited state, n = 4

E₄ = E₁ /4²

E₄ = (1.23 eV) / 16

E₄ = 0.077 eV

Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV

The energy that must be added to the electron to move it to the third excited state is -1.153 eV

For fourth excited state, n = 5

E₅ = E₁ /5²

E₄ = (1.23 eV) / 25

E₄ = 0.049 eV

Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV

The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV

5 0

3 years ago

An object is in circular motion. How will the object behave if the centripetal force is removed

stich3 [128]

The object will sail away in a straight line ... continuing in the same direction it was going when the centripetal force stopped.

3 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

2 years ago

Several light bulbs, each of resistance 1.5 Ω, are connected in a series across a 120 V source of emf. If the current through th

Leni [432]

<h3><u>Answer;</u></h3>

40 light bulbs

<h3><u>Explanation</u>;</h3>

The total resistance of components or bulbs in series is given as the sum of resistance of all the components.

Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ

From the ohms law;

V = IR , where V is the voltage, I is the current and R is the resistor.

Thus; R = V/i

R = 120/2

= 60 Ω

But, there are n bulbs each with 1.5 Ω; thus there are;

n = 60/1.5

<u> = 40 Bulbs </u>

7 0

3 years ago