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bija089 [108]
1 year ago
7

what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el

ectron that is accelerated from rest through a potential difference of 1000 v?
Physics
1 answer:
Evgen [1.6K]1 year ago
5 0

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

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Neglecting air resistance, what maximum height will be reached by an arrow launched straight upward with an initial speed of 35
tankabanditka [31]
The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.

The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)

Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
3 0
3 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

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

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
3 years ago
A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the
sp2606 [1]
From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
F=12N



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

Ft =m(v-u)


12(3.0)=m(6.0- \: 0)
This implies that

36.0 = 6m
m =  \frac{36.0}{6.0}
\therefore \: m = 6.0kg


You can also use the equation of linear motion,
v = u + at
6 = 0 + a(3)
6 = 3a
a =  \frac{6}{3}

a = 2 {ms}^{ - 2}
But
F=ma
12 = m(2)
12 = 2m
\frac{12}{2}  = m
\therefore \: m = 6kg
4 0
3 years ago
Which statement is true of earths poles?
djverab [1.8K]
Answer choice d is correct
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A long wire carries a current toward the north in a magnetic field that is directed vertically downward perpendicular to the sur
xenn [34]

The direction of the magnetic force on the wire is west.

The magnetic force acting on the moving protons acts northward in the horizontal plane. If the thumb is up (current flows vertically up), the wrapped finger will be counterclockwise.

Therefore, the direction of the magnetic field is counterclockwise. Here, the magnetic field is pointing upwards (vertical magnetic field) and the electrons are moving east. Applying Fleming's left-hand rule here, we can see that the direction of force is along the south direction.

As the change in magnetic flux increases upwards, Lenz's law indicates that the induced magnetic field of the induced current must resist and the inside of the loop must be directed downwards. Using the right-hand rule, we can see that a clockwise current is induced.

Learn more about the magnetic fields here: brainly.com/question/7802337

#SPJ4

3 0
2 years ago
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