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3 years ago

An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.

Physics

1 answer:

Minchanka [31]3 years ago

6 0

Answer:

Fₓ = 0, $F_{y}$ = 0 and $F_{z}$ <em> = - 3.115 10⁻¹⁵ N</em>

Explanation:

The magnetic force given by the expression

F = q v xB

the bold are vectors, the easiest analytical way to determine this force in solving the determinant

$F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right] 10^{4}$

F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

F =i^0 + j^0 - k^ 3.115 10⁻¹⁵ N

Fₓ = 0

$F_{y}$ = 0

$F_{z}$ <em> = - 3.115 10⁻¹⁵ N</em>

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Compare a change in temperature of 1°C to a change of 1°F.

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Answer:

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Explanation:

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3 years ago

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A body with a mass of 2,000 kg is lifted to a height of 15 m within a time of 15 s. Which one of the following statements concer

Aleksandr [31]

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3 years ago

You drive on Interstate 10 from San Antonio to Houston, half the time at 72 km/h and the other half at 98 km/h. On the way back

zimovet [89]

Answer: a. 85km/hr b.82.3km/hr

c. 84km/hr

Explanation: first let take the total time from San Antonio to Houston to be 2hr.

Half time 1hr was covered with speed of 72km/hr

Distance = speed*time=72km/hr *1hr

=72km

So too with the second half of 1hr covered with speed of 98km/hr

Distance = 98km

Total distance from Houston to San Antonio is 98+72 =170km

a. Average speed from San Antonio to Houston is

S1 =170/2

=85km/hr

b.half distance from Houston to San Antonio which is 170km/2

= 85km was covered with speed of 72km/hr first half, so time

t = dist/speed

t = 85/72 = 1hr 12 mins

Remaining 85 km covered with a speed of 98km/hr

Time = 85/98 = 0.88*60min

= 52 mins

Total time = 1hr +12mins +52mins

=2hr4mins= 124/60 hr

So average speed = distance/time

=170/124/60

Using reciprocal law

Average speed S2= 170*60/124

= 82.3km/hr

C. Average speed to and fro(entire tripe)

= (85+82.3)/2

=84km/hr

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3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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3 years ago

This chart shows characteristics of three different types of waves.

prohojiy [21]

Answer:

1. Primary or P waves are push and pull waves

2. Secondary, S or Shear Waves are also called transverse wave

3. L or surface waves reach the earth's surface after P and S waves

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3 years ago