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adell [148]
2 years ago
5

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

rt a force on a stationary charged object? yes no (b) does a magnetic field do so? yes no (c) does an electric field exert a force on a moving charged object? yes no (d) does a magnetic field do so? yes no (e) does an electric field exert a force on a straight current-carrying wire? yes no (f) does a magnetic field do so? yes no (g) does an electric field exert a force on a beam of moving electrons? yes no (h) does a magnetic field do so? yes no
Physics
1 answer:
matrenka [14]2 years ago
6 0
<span> (a) does an electric field exert a force on a stationary charged object? 
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
You might be interested in
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
1 year ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
An object moving with a speed of 35M/s and has a kinetic energy of 1500 J what is the mass of the object
Lelu [443]

2.45 Kg

Explanation:

K.E = 0.5 mv^2

1500 = 0.5 × m × (35)^2

1500= m × 0.5 × 1225

1500 = m × 612.5

1500/612.5 = m

2.45 = m

m = 2.45kg

7 0
1 year ago
Odległość między kolejnymi grzbietami fal na morzu wynosi 20 m. Łódź opada z grzbietu fali, unosi się i osiąga ponownie najwyższ
Veronika [31]

Answer:

Explanation:

The distance between successive wave crests at sea is 20 m. The boat descends from the crest of the wave, rises and reaches the highest position again within 5 s. Calculate the wave propagation speed.

Given that,

The distance between two successive crest is 20m

Wavelength is the distance between two successive crest or trough

Then, it's wavelength is λ = 20m

The time to reached the maximum height is 5seconds, then it will take (5×4) to complete one period

Then,

Period T = 20seconds

From wave equation

v = fλ

Where

v is speed

f is frequency and

λ is wavelength

The frequency is related to the period

f =  1 / T

Then,

v = λ / T

So, v = 20 / 20

v = 1 m/s

The speed of propagation of the wave is 1m/s

To Polish

Jeśli się uwzględni,

Odległość między dwoma kolejnymi grzebieniami wynosi 20 m

Długość fali to odległość między dwoma kolejnymi grzebieniami lub dolinami

Zatem jego długość fali wynosi λ = 20 m

Czas do osiągnięcia maksymalnej wysokości wynosi 5 sekund, a następnie ukończenie jednego okresu zajmie (5 × 4)

Następnie,

Okres T = 20 sekund

Z równania falowego

v = fλ

Gdzie

v to prędkość

f oznacza częstotliwość, a

λ jest długością fali

Częstotliwość jest związana z okresem

f = 1 / T

Następnie,

v = λ / T

Zatem v = 20/20

v = 1 m / s

Prędkość propagacji fali wynosi 1m/s

6 0
2 years ago
If a rainstorm drops 5 cm of rain over an area of 13 km2 in the period of 3 hours, what is the momentum (in kg · m/s) of the rai
stiks02 [169]

To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.

The expression that defines the linear momentum is expressed as

P=mv

Where,

m=mass

v= velocity

According to our data we have to

v=10m/s

d=0.05m

A=13*10^6m^2

Volume (V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3

t = 3hours=10800s

\rho = 1000kg/m^3

From the given data we can calculate the volume of rain for 5 seconds

V' = \frac{V}{t}*\Delta t_{total}

Where,

\Delta t_{total} It is the period of time we want to calculate total rainfall, that is

V' = \frac{3.9*10^5}{10800}*5

V' = 1.805*10^2m^3

Through water density we can now calculate the mass that fell during the 5 seconds:

m' = V'*\rho

m' = 1.805*10^2*1000

m' = 1.805*10^5m^2

Now applying the prevailing equation given we have to

P=m'v

P = (1.805*10^5)(10)

P = 1.805*10^6 Kg.m/s

Therefore the momentum of the rain that falls in five seconds is 1.805*10^6 Kg.m/s

4 0
2 years ago
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