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sladkih [1.3K]
1 year ago
9

Suppose we were to attempt to use a similar machine to measure the charge-to-mass ratio of protons, instead. Suppose, for simpli

city, that we can get a "source" for a beam of protons as we do here for electrons (with a similar setup). What complications (if any) do you expect?
Physics
1 answer:
Igoryamba1 year ago
3 0

Answer:

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

Explanation:

The machine to measure the ratio of charge / mass of the electron, has two parts: a part where it accelerates the electrons in an electric field and a second section to charter the beam and measure its radius of curvature calculated from here the q / m ratio

In the case of having protons, the charge has the same value as that of the electrons, but with a positive charge, so the polarities of the fields should be changed.

The mass of the protons is much greater than the mass of the electrons, which introduced a significant difference in the excrement, since similar electric fields the speed of the protons is much less than the speed of the electrons, so the magnetic field through which the voters pass to have equivalent deflations in many cases this small values ​​of the magnetic field are not desirable due to the interference of the Earth's magnetic field.

Another solution could be to increase the electric field to have the protons with speeds similar to the electors, this possibility is not easy either, because the field of trunking of more than 5000 V would be needed, which are very dangerous.

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

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A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the
Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

A) 0.45 mm

B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

Current in the fuse is 1.0 A

Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

d^{2} =\sqrt{ 2.05*10^-7}

d = 4.5 x 10^-4 m = <em>0.45 mm</em>

8 0
1 year ago
Two identical boxes are being pulled across a horizontal floor at a constant velocity by a horizontal pulling force of 176 N tha
MAXImum [283]

Answer:

the tension in the rope between the boxes is equal to 88 N

Explanation:

given,

the force applied on one body F = 176 N

When two bodies are moving on horizontal plane at constant velocity then their kinetic friction (f k) is equal to applied force F

According to newton third law the resultant force acting on one body is equal to the resultant force acting on the another body.

T is the tension in the rope

T- f_k = - (T- f_k)

T - F = - (T - F)

T - 176 = - (T - 0)

2 T = 176

T = 176/2 = 88 N

so, the tension in the rope between the boxes is equal to 88 N

3 0
2 years ago
Read 2 more answers
In a charge-free region of space, a closed container is placed in an electric field. Which of the following is a requirement for
galina1969 [7]

Answer:

D. The requirement does not exist -the total electric flux is zero no matter what.

Explanation:

According to Gauss's law , total electric flux over a closed surface is equal to 1 / ε₀ times charge inside.

If charge inside is zero , total electric flux over a closed surface is equal to

zero . It has nothing to do with whether external field is uniform or not. For any external field , lines entering surface will be equal to flux going out.

8 0
2 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0
2 years ago
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