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pashok25 [27]
3 years ago
12

An ion source is producing 6Li ions, which have charge +e and mass 9.99 × 10-27 kg. The ions are accelerated by a potential diff

erence of 13 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.0 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.
Physics
1 answer:
Ivanshal [37]3 years ago
8 0

Answer:

6.45 x 10^5 N/C

Explanation:

q = + e = 1.6 x 10^-19 C

m = 9.99 x 10^-27 kg

V = 13 kV = 13000 V

B = 1 T

Let v be the speed and E be the strength of electric field.

1/2 mv^2 = eV

v^2 = 2 e v / m

v^2 = (2 x 1.6 x 10^-19 x 13000) / (9.99 x 10^-27)

v = 6.45 x 10^5 m/s

As the charge particle is undeflected, the force due to magnetic field is counter balanced by the force due to electric field.

q E = q v B

E = v B = 6.45 x 10^5 x 1 = 6.45 x 10^5 N/C

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Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?
hichkok12 [17]

Answer:

t=5.36\times 10^{-6}\ s

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s

So, the required time is 5.36\times 10^{-6}\ s

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An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
3 years ago
What is SI unit for distance
Dafna11 [192]

Answer:

meter

The SI unit of distance and displacement is the meter [m].

Explanation:

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3 0
4 years ago
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Which sentence states Newton’s third law?
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<span>If two objects collide, each object exerts a force equal to and in the opposite direction of the other.</span>
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Additional Problem: A simple pendulum, consisting of a string (of negligible mass) of length L with a small mass m at the end, i
kykrilka [37]

Answer:

a)   v = √ 2gL  abd  b)  θ = 45º

Explanation:

a) for this part we use the law of conservation of energy,

Highest starting point

       Em₀ = U = mg h

Final point. Lower

       Em₂ = ½ m v²

      Em₀ = Em₂

      m g h = ½ m v²

      v = √2g h

      v = √ 2gL

b) the definition of power is the relationship between work and time, but work is the product of force by displacement

     P = W / t = F. d ​​/ t = F. v

If we use Newton's second law, with one axis of the tangential reference system to the trajectory and the other perpendicular, in the direction of the rope, the only force we have to break down is the weight

     sin θ = Wt / W

     Wt = W sin θ

This force is parallel to the movement and also to the speed, whereby the scalar product is reduced to the ordinary product

     P = F v

The equation that describes the pendulum's motion is

    θ = θ₀ cos (wt)

Let's replace

    P = (W sin θ) θ₀ cos (wt)

    P = W θ₀ sint θ cos (wt)

We use the equation of rotational kinematics

    θ = wt

    P = Wθ₀ sin θ cos θ

Let's use

   sin 2θ = 2 sin θ cos θ

   P = Wθ₀/2 sin 2θ

This expression is maximum when the sine has a value of one (sin 2θ = 1), which occurs for 90º,

    2θ = 90

    θ = 45º

5 0
3 years ago
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