What is the shape of the trajectory that a charged particle follows in a uniform magnetic field?

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

4 years ago

The earth’s magnetic field is located in the _____.

Natali [406]

Near Greenland in the northern hemisphere <span />

7 0

3 years ago

With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor

Fofino [41]

Answer:

A) K / K₀ = 4 b) v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

½ k 4 x₀² = K

(½ K x₀²) = K₀

K = 4 K₀

K / K₀ = 4

B) the speed value

½ k 4 x₀² = ½ m v²

v = 4 (k / m) x₀

if we call

v₀ = k / m x₀

v = 4 v₀

v / v₀ = 4

3 0

3 years ago

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

or

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.

Solving for $v$ we get:

$v \geq \sqrt{\dfrac{2GM}{R} }$

putting in numerical values gives

$v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }$

$\boxed{v\geq 5739.5m/s}$

in kilometers this is

$v\geq5.7395m/s.$

Hence, the minimum speed required is 5.7395km/s.

5 0

3 years ago

Answer all please!!!!!!!!!!✌

Nina [5.8K]

Answer:

koj ruam zog

Explanation:

7 0

3 years ago

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