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pishuonlain [190]
3 years ago
10

Heavy ions, such as alpha particles, lose kinetic energy as they travel through matter. Consider equation 31.1 or 31.2. Where do

heavy ion lose more kinetic energy?
a. The rate at which they lose energy does not depend on the speed of the ion.
b. When they are traveling at high speeds, just at they enter a medium
c. When they are traveling at low speeds, when they are almost stopped
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

c. When they are traveling at low speeds, when they are almost stopped

Explanation:

The rate at which they ions loose energy depends upon its charge and mass and not on its speed. For example, We know that alpha particles have high mass and high charge so they loose more energy than gamma radiation which has no mass and no charge. The slower heavy particle which has high linear energy transfer.

Hence, when they travel at low speed they are almost stopped.

There the correct answer is option C

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A wave travels at 175 m/s along the x-axis.If the period of the periodic vibrations of the wave is 3.00 milliseconds,then what i
OlgaM077 [116]

Answer:  

<h2>E) 52.5 cm</h2>

Explanation:

Step one:

given data

period T= 3 milliseconds= 0.003

velocity v= 175m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=175*0.002

λ=0.525m

to cm= 0.525*100

=52.5cm

The wavelength of the wave is E) 52.5 cm

7 0
2 years ago
Please help on this one?
Nonamiya [84]

30 or c because 60-30=30

3 0
3 years ago
Two point charges of equal magnitude are 8.0 cm apart. At the midpoint of the line connecting them, their combined electric fiel
bagirrra123 [75]

Answer:

r = 8/2 = 4cm = 0.04m

k = 9×10^9

Enet = 51 N/C

Enet = E1 + E2

since E1 = E2

E1 = Enet/2 = 51/2

E/2 = kq/r²

q = Er²/2k

q = (51 × 0.04²)/(2×9×10^9)

q = 4.5×10^-12 C

q1 = q2 = 4.5 pC

Explanation:

The electric field is a region around a

charge in which it exerts electrostatic force

on another charges. While the strength of

electric field at any point in space is called

electric field intensity. It is a vector

quantity. Its unit is NC¯¹.

According to coulomb’s law ,if a unit

positive charge q (call it a test charge) is

brought near a charge q (call a field

charge) placed in space,the charge q will

experience a force. The value of this force

depends upon the distance between the

two charges. If the charge q is moved

away from q ,this force would decrease till

at a certain distance the force would be

practically reduced to zero. The charge q

is then out of the influence of charge q.

The region of space surrounding the charge

q in which it exerts a force on the charge

q is known as E.F of the charge

q. Mathematically it is expressed as:

E =F/q

The direction of the vector E is the same

as the direction of F,because q is a

positive scalar. Dimensionally,the E.F is

force per unit charge,and its SI unit is the

newton/coulomb (N/C).

7 0
2 years ago
In phase E, from what direction is the sun shining?
Zanzabum

Heyo,

Your answer is West

if this question helped can you mark me brainliest

8 0
3 years ago
Read 2 more answers
A uniform magnetic field of 4.15 T points in some direction. Consider the magnetic flux through a large triangular wire loop tha
pickupchik [31]

Answer:

ФB = 4.89 W

Explanation:

In order to calculate the magnetic flux trough the triangular wire loop, you use the following formula:

\Phi_B=B\cdot A=BAcos\theta   (1)

B: magnitude of the magnetic field = 4.15T

A: area of the triangular loop

θ: angle between the normal to the surface and the direction of the magnetic field = 0°

Both direction of the magnetic field and the normal vector to the surface are parallel.

You calculate the area of the triangular loop:

A=\frac{bh}{2}   (2)

b: base = 1.65 m

h: height = \sqrt{(1.65m)^2-(\frac{1.65m}{2})^2}=1.42m

A=\frac{(1.65m)(1.42m)}{2}=1.17m^2

Next, you replace the values of A and B in the equation (1):

\Phi_B=(4.15T)(1.17m^2)=4.89W

The magnetic flux trough the triangular wire loop is 4.89W

7 0
3 years ago
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