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2 years ago

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An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magn

etic field at the following (x, y, z) positions?
a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 1 cm )
c. (0 cm, 2 cm , 1 cm )

1 answer:

Hitman42 [59]2 years ago

5 0

This answer of the strength and directions of the magnetic field is a

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It feels better to catch a hard ball while wearing a padded glove than to catch it bare handed. use the ideas of momentum and im

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Answer:

Let's say the pitcher is angry or just has a really heavy hand while throwing this ball, and now you have to catch it, otherwise it's going to ram into your face. When you put your hands up just in time to catch this ball, this is called impulse, or commonly expressed as a reflex. Depending on what kind of ball is being thrown, such as a golf ball, baseball, basketball, beach-ball, rubber-ball, baseball, etc. ... the weight of the ball itself is going to impact how much it i going to hurt when you catch it without any hand protection. However, if you're catching, let's say a baseball, with a padded glove, it is not going to hurt as bad as catching the baseball bare handed, because the padded glove has enough padding in it to create a barrier between the skin of your hand and the palm of the glove.

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3 years ago

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif

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Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

$v(t)=-gt-v_e\times \ln \frac{m-rt}{m}$

v = velocity of rocket at time t

g = Acceleration due to gravity = $9.8 m/s^2$

$v_e$ = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

$v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})$

$v(60) = 668.97 m/s$

Height of the rocket = h

$Velocity=\frac{Displacement}{time}$

$668.97 m/s=\frac{h}{60 s}$

$h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km$

Height of the rocket be one minute after liftoff is 40.1382 km.

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2 years ago

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Answer and Explanation:

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(b). The electron will follow a circular path which travels along under constant magnetic field

$r = \frac{mV}{qB}$

where m = mass of electron, V = velocity of electron, q = charge of the electron and B = magnetic field strength

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