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Mazyrski [523]
2 years ago
7

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz =

−4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?
Physics
1 answer:
cestrela7 [59]2 years ago
3 0

Explanation:

It is given that,

Charge on the particle, q=5\ nC=5\times 10^{-9}\ C

Mass of the particle, m=3\ \mu g=3\times 10^{-6}\ g=3\times 10^{-9}\ kg

Magnetic field component, B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT

Net magnetic field, B=\sqrt{2^2+3^2+4^2}=5.38\ mT=0.00538\ T

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, \theta=120

Magnetic force is given by :

F=qvB\ sin\theta

F=5\times 10^{-9}\times 5000\ m/s\times 0.00538\times sin(120)

F=1.16\times 10^{-7}\ N

Acceleration of the particle is given by, a=\dfrac{F}{m}

a=\dfrac{1.16\times 10^{-7}\ N}{3\times 10^{-9}\ kg}

a=38.6\ m/s^2

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.

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If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

7 0
2 years ago
Read 2 more answers
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bagirrra123 [75]

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

Explanation :

Objects have three subatomic particles that are Electrons, protons, and neutrons.

Protons and neutrons are found in the nucleus and electrons rotate or move outside the nucleus. Naturally, protons are positively charged, neutrons have no charge, and electrons are negatively charged.

Therefore, an object that is negatively charged has more electrons than protons.  An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

8 0
2 years ago
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Murljashka [212]

Answer:

When an object changes speed (increases/decreases) it results in acceleration/de acceleration, its velocity also changes.

Explanation:

Acceleration is the rate of change in velocity.An object can accelerate when speed increases, decreases or direction changes. All these instances involves a change in velocity.Velocity is a vector quantity thus it has magnitude and the direction.Acceleration due to change in direction is centripetal acceleration.The expression for finding acceleration is;

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3 0
2 years ago
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aksik [14]
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2 years ago
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