Of the following metals, which is the MOST reactive? A. Li B. Be C. Na D. Mg

An electron in a vacuum is first accelerated by a voltage of 51400 V and then enters a region in which there is a uniform magnet

zimovet [89]

Answer:

F = 8.6 10⁻¹² N

Explanation:

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Em₀ = U = q ΔV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v²

Em₀ = Emf

e ΔV = ½ m v²

v =√ 2 e ΔV / m

v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)

v = √(1.8075 10¹⁶)

v = 1,344 10⁸ m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4

For this exercise we use the law of conservation of energy

Initial. Field energy with the electron at rest

Emo = U = q DV

Final. Electron with velocity, just out of the electric field

Emf = K = ½ m v2

Emo = Emf

.e DV = ½ m v2

.v = RA 2 e DV / m

.v = RA (2 1.6 10-19 51400 / 9.1 10-31)

.v = RA (1.8075 10 16)

.v = 1,344 108 m / s

Now we can use the equation of the magnetic force

F = q v x B

Since the speed and the magnetic field are perpendicular the force that

F = e v B

F = 1.6 10-19 1,344 108 0.4

F = 8.6 10-12 N

5 0

3 years ago

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i

ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

$m_1v_1+m_2v_2=(m_1+m_2)v$

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

$m_1(-v_1)+m_2v_2=(m1+m2)(-v2)$

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

$-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s$

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

$v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s$

hence, the time is t=0.42 s

4 0

3 years ago

A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.

zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = $7.50\times 10^{- 3}\ H$

Current in the coil, i = $1680cos[\frac{\pi t}{0.0250}]$ A

where

$i_{max} = 1680\ mA = 1.680\ A$

Now,

(a) To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

$e = \frac{Ldi}{dt}$

$e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]$

$e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]$

For maximum emf, $sin\theta$ should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

$|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V$

(b) To calculate the maximum average flux,we know that:

$\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb$

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

$e = e_{o}sin{\pi t}{0.0250}$

$e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V$

7 0

3 years ago

Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp

shutvik [7]

Answer: 500 Watts

Explanation:

Power $P$ is the speed with which work $W$ is done. Its unit is Watts ( $W$ ), being $1 W=\frac{1 Joule}{1 s}$ .

Power is mathematically expressed as:

$P=\frac{W}{t}$ (1)

Where $t$ is the time during which work $W$ is performed.

On the other hand, the Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path. It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter ( $1J=(1N)(1m)=Nm$ ).