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3 years ago

PLEASE HELP URGENT

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At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

8 0

3 years ago

A 0.0223 m diameter coin rolls up a 12.0◦ inclined plane. The coin starts with an initial angular speed of 49.3 rad/s and rolls

zlopas [31]

Answer:

$h = 0.0231 m$

Explanation:

The movement of the coin is modelled after the Principle of Energy Conservation. The kinetic energy of the coin is the sum of the components associated with translation and rotation and there are no non-conservative forces. In addition, the coin starts moving at height of zero

$K_{rot} + K_{tr} = U_{g}$

$\frac{1}{2} \cdot m_{coin} \cdot \omega^{2} \cdot R^{2} + \frac{1}{4} \cdot m_{coin} \cdot R^{2} \cdot \omega^{2} = m_{coin} \cdot g \cdot h\\$

$\frac{3}{4} \cdot R^{2} \cdot \omega^{2} = g \cdot h$

The maximum vertical height is isolated in the previous equation:

$h = \frac{3 \cdot R^{2}\cdot \omega^{2}}{4\cdot g}$

$h = \frac{3 \cdot (0.01115 m)^{2} \cdot (49.3 \frac{rad}{s} )^{2}}{4 \cdot (9.807 \frac{m}{s^{2}} )} \\h = 0.0231 m$

5 0

4 years ago

I have 510 missing assignments on edg, Any tips on catching up?

RSB [31]

Answer:

Take notes

Explanation:

Cuz it helps you stay on task

6 0

3 years ago

Read 2 more answers

In the oscillating spring ball system, where is the velocity of the ball the greatest?

nlexa [21]

The speed of an object in a mass-spring system is given under the function

$v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}$

Here,

m = mass

k = Spring constant

A = Amplitude

x = Position

When the position is at the equilibrium point (x = 0), the speed will be maximum, and could even be expressed as

$v_{max}= A\sqrt{\frac{x}{m}}$

So the correct answer is B.

6 0

3 years ago

What is the final velocity of the ball that is dropped from a height of 200m?

Hoochie [10]

The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .

<h3>What is velocity with example?</h3>

The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.

<h3>Briefing:</h3>

Given the initial velocity of the ball (u) = 0

Distance travelled by the ball (s) = 200m

Acceleration (a) = 10 m/s²

As we know:

v² = u² + 2as

Putting values:

v² = 0+2 × (10 m/s²) × (200 m)

v = 44.73 m/s.

To know more about Velocity visit:

brainly.com/question/18084516

#SPJ9

8 0

1 year ago