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3 years ago

8

A crew team rows a boat at a rate of 20 km/h in still water. In practice on a river, the team rows for 30 minutes up the river (

against the current), and then for 30 minutes down the river (with the current). The speed of the river current is 1.5 km/h. How much farther did they travel in the second 30 minutes?

1 answer:

sweet-ann [11.9K]3 years ago

6 0

In the second 30 mins, the speed should be 20 + 1.5 = 21.5 km/h

So S = 21.5 * 30/60 = 10.75 km

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3 years ago

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Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

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Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

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now plug in all data into above equation

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so correct answer is

A) - 1.8 m/s

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3 years ago

Use the equation for magnetic force on a moving charge to derive the equation for magnetic force on a current carrying wire. Sho

max2010maxim [7]

Answer:

The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F is force and is a vector because the force acts in a direction. q is the charge of the particle. v is velocity and is a vector because the particle is moving in some direction. B is the magnetic flux density.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd. Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore

∮B dl=μ0 I

B ∮dl=μ0 I

B 2πr=μ0 I

B=μ02πr Id=μ0/4π I dl×rr3

Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get

dl×r=−z dlj^+rsinθk^

and therefore the magnitude of dB is equal to

dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3

Thus, magnetic field is depending on r,θ,z.

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1 year ago

In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed

Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is $87.87 * 10^{-37}$ .

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= $5.29 * 10^{-11} m\\$

velocity= $2.9* 10^{6} m/s$

Working formula, M=N/A

$I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}$

$T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }$

= $15.16 * 10^{-5} s$

$I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }$ $= 0.10 * 10^{-14}$

= $1 * 10^{-15} C$

M= $1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}$

= $87.87 * 10^{-37}$

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1 year ago

Show that (a)KE=1/2mv2

evablogger [386]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}$

Explanation:

$\underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}$

Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :

$\sf{W = FD}$

⇒ $\sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)$

⇒ $\sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}$

⇒ $\sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }$

⇒ $\sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}$

The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²

∴ $\sf{KE= \frac{1}{2} m {v}^{2} }$

$\sf{ \underline{ \bold{ {proved}}}}$

Hope I helped!

Best regards!!

5 0

3 years ago