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Ganezh [65]
3 years ago
8

An X-ray tube emits X-rays with a wavelength of 1.0 x 10-11 m. Calculate the potential that must be applied across the X-ray tub

e to give each incident electron its energy, in volts.
A ) 0.6 x 105
B ) 1.2 x 105
C ) 2.4 x 105
D ) cannot be determined
Physics
2 answers:
sergey [27]3 years ago
3 0

Answer:

1.24 x 10 to the 5 ev = 124,000 ev          its B

Explanation:

E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev

h = Planck's constant = 6.626 × 10 to the -34 joule·s

c = speed of light = 2.998 × 10 to the 8 m/s

lambda is the given wavelength

E is the desired photon energy

mel-nik [20]3 years ago
3 0

The potential that must be applied across the X-ray tube to give each incident electron the appropriate energy to goose X-rays with a wavelength of 1.0 x 10-11 m out of the anode cannot be determined from the given information. <em>(D)</em>

The energy (or wavelength) of the emitted X-rays is NOT directly determined by the energy of the incident electrons in the x-ray tube. It's determined mainly by the substance of which the anode is composed, and the electron configuration of the substance.

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Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet
Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

A=\pi r^2

Put the value into the formula

A=\pi\times(5\times10^{-2})^2

A=7.85\times10^{-3}\ m^2

We need to calculate the capacitor

Using formula of capacitor

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}

C=6.94\times10^{-12}\ F

We need to calculate the resistance of the wire

Using formula of resistivity

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}

R=4.27\times10^{-3}\ \Omega

We need to calculate the voltage

Using formula of charge

q=CV

V=\dfrac{q}{C}

Put the value into the formula

V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}

V=1.801\times10^{3}\ V

(a). We need to calculate the current

Using formula of current

I=\dfrac{V}{R}

I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}

I=421779.85\ A

I=4.217\times10^{5}\ A

(b). We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}

E=11.2\times10^{5}\ N/C

The electric field in the wire is 11.2\times10^{5}\ N/C.

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

E=\dfrac{1}{2}CV^2

Put the value into the formula

E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2

E=1.126\times10^{-5}\ J

The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Hence, (a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

8 0
3 years ago
The starship Enterprise approaches the planet Risa at a speed of 0.8c relative to the planet. On the way, it overtakes the inter
Alchen [17]

Answer:

<em>0.3c</em>

<em></em>

Explanation:

The speed of Enterprise relative to Risa is 0.8c

Relative speed of both ships as measured from Enterprise is 0.5c

therefore, relative speed of Astra to Enterprise is 0.8c - 0.5c = <em>0.3c</em>

<em>this is also the relative speed with which Astra approaches the planet Risa since Enterprise's speed was calculated relative to Risa.</em>

8 0
2 years ago
Which item is not considered electromagnetic energy?
suter [353]
Sound waves are known to be the one that's not considered as a type of electromagnetic energy. As for microwaves and x-rays, they tend to share the same frequencies that can be considered as electromagnetic, and sound waves have a different frequency than them.
5 0
3 years ago
A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re
stepan [7]
Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
 (20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
 (20 m/s)*sin(37°) = 12.036 m/s

The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by 
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy  = 12.036 - 9.8*0.2815
   = 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is 
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m  is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is 
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

6 0
2 years ago
A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle:___
iVinArrow [24]

A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle increases

The most common charge carriers are the positively charged proton and the negatively charged electron. The movement of any of these charged particles constitutes an electric current

<h3>What is a Charge ?</h3>

When there are more or fewer protons in an atom than electrons, the substance has an electric charge. Protons have a positive charge, while electrons have a negative charge. If a substance has more protons than electrons, it is positively charged; if it has more electrons, it is negatively charged.

  • The SI units for charge are ampere-second or coulomb. When one ampere of electric current goes through the conductor for one second, one coulomb of charge passes through it. Charge is denoted by the formula Q = I t.

Learn more about Charge here:

brainly.com/question/18102056

#SPJ4

7 0
1 year ago
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