1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Charra [1.4K]
3 years ago
9

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

are 0.00 cm and 2.30 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0370 T. Determine the energy (in keV) of the incident electron.
Physics
1 answer:
morpeh [17]3 years ago
3 0

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron,u_2=0

Radius,r_1=0

r_2=2.3 cm=\frac{2.3}{100}=0.023 m

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron,m=9.1\times 10^{-31} kg

Charge on an electron,q=-1.6\times 10^{-19} C

Velocity,v=\frac{Bqr}{m}

Using the formula

Speed of electron,v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0

Speed of second electron,v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}

v_2=1.5\times 10^8 m/s

Kinetic energy of incident electron=\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2

Kinetic energy of incident electron=0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J

Kinetic energy of incident electron=\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV

1KeV=1000eV

You might be interested in
Why potential energy become equal to kinetic energy at height
Gennadij [26K]

Answer:

because potentil energy is redy to go but its bound up

And kinetic energy is in motion

Explanation:

7 0
3 years ago
Read 2 more answers
A material you are testing conducts electricity but canot be pulled into wires
agasfer [191]
A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid. Hope this helps!
4 0
3 years ago
4. I drop a pufferfish of mass 5 kg from a height of 5.5 m onto an upright spring of total length 0.5 m and spring constant 3000
KatRina [158]

Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

Now by using the conservation of energy; we realize that :

Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

Replacing our given values into the above equation; we have :

(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

269.5 = 47.5 x + 1500(0.5 -x )²

269.5 = 47.5 x + 1500(0.25 - x²)

269.5 = 47.5 x + 375 - 1500 x²

269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

8 0
3 years ago
newton's second law states that when a net force acts on an object, it accelerates it.Explain how it would be possible for two o
Alika [10]

Answer:

See Explanation

Explanation:

According to Newton's second law of motion, the acceleration of a body is proportional to the net external force that acts on the body.

A body accelerated when it is acted upon by an unbalanced net external force.

When the external forces acting on a body are balanced, the effect of each force is cancelled by the other hence the body is not accelerated according to Newton's second law.

4 0
2 years ago
A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The
Yuki888 [10]

Answer:

Approximately 0.560\; {\rm m}, assuming that:

  • the height of 3.21\; {\rm m} refers to the distance between the clay and the top of the uncompressed spring.
  • air resistance on the clay sphere is negligible,
  • the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}, and
  • the clay sphere did not deform.

Explanation:

Notations:

  • Let k denote the spring constant of the spring.
  • Let m denote the mass of the clay sphere.
  • Let v denote the initial speed of the spring.
  • Let g denote the gravitational field strength.
  • Let h denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let x denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of x, the elastic potential energy \text{PE}_{\text{spring}} in this spring would be:

\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}.

The initial kinetic energy \text{KE} of the clay sphere was:

\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}.

When the spring is at the maximum compression:

  • The clay sphere would be right on top of the spring.
  • The top of the spring would be below the original position (when the spring was uncompressed) by x.
  • The initial position of the clay sphere, however, is above the original position of the top of the spring by h = 3.21\; {\rm m}.

Thus, the initial position of the clay sphere (h = 3.21\; {\rm m} above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by (h + x).

The gravitational potential energy involved would be:

\text{GPE} = m\, g\, (h + x).

No mechanical energy would be lost under the assumptions listed above. Thus:

\text{PE}_\text{spring} = \text{KE} + \text{GPE}.

\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x).

Rearrange this equation to obtain a quadratic equation about the only unknown, x:

\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0.

Substitute in k = 1570\; {\rm N \cdot m^{-1}}, m = 4.87\; {\rm kg}, v = 5.21\; {\rm m\cdot s^{-1}}, g = 9.81\; {\rm m \cdot s^{-2}}, and h = 3.21\; {\rm m}. Let the unit of x be meters.

785\, x^{2} - 47.775\, x - 219.453 \approx 0 (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that x \ge 0:

\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}.

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately 0.560\; {\rm m}.

5 0
2 years ago
Other questions:
  • Scientific question must be
    9·1 answer
  • What is the relationship between energy use and motion?
    5·1 answer
  • Is nickel coin magnetic?
    12·2 answers
  • 1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane
    13·1 answer
  • How much work is done if a 4500 N piano is lifted 10 m?
    13·1 answer
  • Help plsssssssssss I write it 100 time no one answer
    15·1 answer
  • What is the equivalent resistance of the circuit pictured below?
    7·2 answers
  • What important example, shown in the picture, is given regarding a real-life
    5·1 answer
  • 7.a railway truck A of a mass of 2000kg moves westwards with a velocity of 3m/s. It collides with a stationary truck B 1200kg, l
    5·1 answer
  • The passengers are facing forward in the direction that the train is moving. What will happen to the passengers sitting on a tra
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!