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3 years ago

Which of the following are for vector directione?

1 answer:

8 0

Answer:
C. outside 45 degrees
Step-by-step explanation:
C. outside 45 degrees
because it can be any point or direction outside 45 degrees

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Describe the motionof an object that gas acceleration of 0 m/s

Maru [420]

If an object has acceleration of zero meters per second², then the object is traveling at a constant speed.

Note: The constant speed doesn't have to be zero, so the object is not necessarily at rest. It can be moving at any speed, as long as the speed doesn't change. (If the speed changed, then the acceleration wouldn't be zero.)

3 0

4 years ago

Afina-wow [57]

Answer:

2.76×10⁻¹⁰ C

Explanation:

Applying,

V = W/q................... Equation 1

Where V = Electric Potential, E = Electric potential energy, q = charge.

make q the subject of the equation

q = W/V................ Equation 2

From the question,

Given: W = 4.26×10⁻⁸ J, V = 154.5 V

Substitute these values into equation 2

q = 4.26×10⁻⁸/154.5

q = 2.76×10⁻¹⁰ C

3 0

3 years ago

Wat's the definition of microwave spectroscopy

ivann1987 [24]

Answer:

Microwave spectroscopy is the spectroscopy method that employs microwaves, i.e. electromagnetic radiation at GHz frequencies, for the study of matter

6 0

4 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

4 years ago

What formed as a result of a hot spot?

masha68 [24]

A the San Andreas fault

7 0

4 years ago

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