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An electron experiences a magnetic force of magnitude 4.60 x 10^-15 N when moving at an angle of 60.0° with respect to a magnetic field of magnitude 3.50 x 10^-3 T. Find the speed of the electron.

Answers

Answer 1

Answer:

Answer:

9.49 × 10⁶ m/s

Explanation:

Data provided in the question:

Magnitude of Magnetic force, F = 4.60 × 10⁻¹⁵ N

Angle, θ = 60°

Magnitude of magnetic field, B = 3.50 × 10⁻³ T

Now,

we know

F = qVBsin(θ)

here ,

q is the charge of electron = 1.6 × 10⁻¹⁹ V

V is the speed of electron

Therefore,

4.60 × 10⁻¹⁵ = ( 1.6 × 10⁻¹⁹ ) × V × ( 3.50 × 10⁻³ ) × sin(60°)

V = 9.49 × 10⁶ m/s

Answer 2

Answer:

Answer:

9.49*10^6(m/s)

Explanation:

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The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

A radioactive nucleus has a half-life of 5*108 years. Assuming that a sample of rock (say, in an asteroid) solidified right after the solar system formed, approximately what fraction of the radioactive element should be left in the rock today? The age of the solar system is 4.5*109 years. (No calculator is necessary, but if it helps use it.)

Answers

Answer:

0.002

Explanation:

The half-life of the radioactive nucleus is related to its quantity, by the following equation:

$N_(t) = N_(0)2^{-t/t_(1/2)}$

Where:

N(t): is the quantity of the radioactive nucleus at time t

N₀: is the initial quantity of the radioactive nucleus

t: is the time = 4.5x10⁹ years

t(1/2): is the half-life of the radioactive nucleus = 5x10⁸ years

$(N_(t))/(N_(0)) = 2^{-4.5 \cdot 10^(9) y/5 \cdot 10^(8) y} = 2 \cdot 10^(-3)$

Therefore, the fraction of the radioactive element in the rock today is 0.002.

I hope it helps you!

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.

Answers

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

$\Sigma F = F - m\cdot g = m\cdot a$ (1)

Where:

$F$ - Buoyant force, measured in newtons.

$m$ - Mass of the plastic ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration, measured in meters per square second.

If we know that $F = 5\,N$ , $m = 0.408\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then the net acceleration of the plastic ball is:

$a = (F)/(m) - g$

$a= 2.448\,(m)/(s^(2))$

The acceleration is 2.448 meters per square second and is vertically upward.

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?

Answers

The acceleration of the particle is 38.87 kg.

Net magnetic field

The net magnetic field is calculated as follows;

$B_(net) = √(B_x^2 + B_y^2 + B_z^2) \n\nB_(net) = √(2^2 + 3^2 + 4^2) = 5.385 \ mT$

Magnetic force on the charge

The magnetic force on the charge is calculated as follows;

$F = qvB * sin(\theta)\n\nF = 5* 10^(-9) * 5* 10^3 * 5.385 * 10^(-3) * sin(120)\n\nF = 1.166 * 10^(-7) \ N$

Acceleration of the particle

The acceleration of the particle is calculated as follows;

$a = (F)/(m) \n\na = (1.166 * 10^(-7))/(3 * 10^(-9)) \n\na = 38.87 \ kg$

Learn more about magnetic force here: brainly.com/question/13277365

Explanation:

It is given that,

Charge on the particle, $q=5\ nC=5* 10^(-9)\ C$

Mass of the particle, $m=3\ \mu g=3* 10^(-6)\ g=3* 10^(-9)\ kg$

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

Net magnetic field, $B=√(2^2+3^2+4^2)=5.38\ mT=0.00538\ T$

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

Angle between velocity and magnetic field, $\theta=120$

Magnetic force is given by :

$F=qvB\ sin\theta$

$F=5* 10^(-9)* 5000\ m/s* 0.00538* sin(120)$

$F=1.16* 10^(-7)\ N$

Acceleration of the particle is given by, $a=(F)/(m)$

$a=(1.16* 10^(-7)\ N)/(3* 10^(-9)\ kg)$

$a=38.6\ m/s^2$

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

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?