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Crazy boy [7]
3 years ago
6

A proton traveling to the right enters a region of uniform magnetic field that points into the screen. When the proton enters th

is region, it will be..??
deflected toward bottom of the screen
deflected out of the plane of screen
deflected toward top of the screen
deflected into plane of screen
unaffected in its direction of motion
Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

deflected toward top of the screen

Explanation:

We can find the direction of the force acting on the proton by using the right-hand rule. In fact, we have:

- Index finger: direction of motion of the proton --> to the right

- Middle finger: direction of the magnetic field --> into the screen

- Thumb: it gives the direction of the force --> therefore, it will be upward

And since the charge is positive (it is a proton), we don't have to revers the direction: so, the force on the proton points upward.

Therefore, the correct answer will be

deflected toward top of the screen

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What is a big five trait
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Answer:

Big 5 Traits are openness conscientiousness, extraversion, agreeableness, and neuroticism.

8 0
3 years ago
A projectile is launched vertically from the surface of the Moon with an initial speed of 1360 m/s. At what altitude is the proj
LenaWriter [7]

Answer:

485520 m

Explanation:

v_{o} = initial velocity of the projectile = 1360 m/s

v_{f} = final velocity of the projectile = \left ( \frac{2}{5} \right )v_{_{o}} = \left ( \frac{2}{5} \right )(1360) = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a h

Inserting the values

544^{2} = 1360^{2} + 2 (-1.6) h

h = 485520 m

3 0
3 years ago
What is the station's orbital speed? the radius of earth is 6.37×106m, its mass is 5.98×1024kg.
Paha777 [63]

Answer:

v_o=2503.08\frac{m}{s}

Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

v_o=\sqrt{\frac{GM}{r}}

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

v_o=\sqrt{\frac{6.674*10^{-11}\frac{m^3}{kg\cdot s^2}(5.98*10^{24}kg)}{6.37*10^6m}}\\v_o=2503.08\frac{m}{s}

3 0
3 years ago
Read 2 more answers
What is shot-curciting​
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Answer:

A path that allows most of the current in an electric circuit to flow around or away from the principal elements or devices in the circuit.

3 0
3 years ago
The displacement of a wave traveling in the positive x-direction is y(x, t)|= (3.5 cm)cos(2.7x − 92t), where x is in m and t is
zubka84 [21]

Answer

Given,

y(x, t) = (3.5 cm) cos(2.7 x − 92 t)

comparing the given equation with general equation

y(x,t) = A cos(k x - ω t)

 A = 3.5 cm  , k = 2.7 rad/m    , ω = 92 rad/s

we know,

a) ω =2πf

   f = 92/ 2π

   f = 14.64 Hz

b) Wavelength of the wave

 we now, k = 2π/λ

      2π/λ = 2.7

      λ = 2 π/2.7

      λ = 2.33 m

c) Speed of wave

     v = ν λ

     v = 14.64 x 2.33

     v = 34.11 m/s

5 0
3 years ago
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