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3 years ago

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An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

e electric force on this electron due to this field?
1.6 × 10-3 newtons
1.4 × 1024 newtons
2.2 × 10-14 newtons
7.4 × 10-13 newtons
4.5 × 1014 newtons

2 answers:

RSB [31]3 years ago

7 0

F=Eq. Just sub in the values and you are done.

ivann1987 [24]3 years ago

3 0

Answer:

2.2 x 10^-14 N

Explanation:

q = - 1.6 x 10^-19 C

E = 1.4 x 10^5 N/C

The force on a charged particle placed in an electric field is given by

F = q x E

F = - 1.6 x 10^-19 x 1.4 x 10^5 = - 2.24 x 10^-14 N

F = - 2.2 x 10^-14 N

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lina2011 [118]

The answer is C. You must divide your wavelength and your frequency to get your answer.

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3 years ago

A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the

meriva

Answer:

a. $2.86 m/s^2$

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

$\mu_s=0.14$

$g=9.8 m/s^2$

By Newton's second law

$8F-f=(8M+m)a$

$a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}$

$a=2.86 m/s^2$

b.By Newton's second law

$T=ma+\mu_s mg$

Substitute the values

$T=250\times 2.86+0.14(250)(9.8)=1058 N$

Hence, the force in the coupling between the dogs and the sled=1058 N

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3 years ago

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Juliette [100K]

Answer:

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(I searched it up since I never learned this)

Explanation:

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3 years ago

An ideal ammeter would have zero resistance while an ideal voltmeter would have infinite resistance, why?.

Dafna1 [17]

Answer: An ideal ammeter would have zero resistance, because to ensure that, there is no voltage drop due to the internal resistance. Similarly, an ideal voltmeter would have infinite resistance, because to ensure that there is no current is drawn by the voltmeter.

Explanation: To find the answer, we need to know about the Ammeter and Voltmeter.

<h3>What is an ammeter?</h3>

An ammeter is a device, that can be used to measure the electric current flows through a circuit in amperes.

An ideal ammeter would have zero resistance, because to ensure that, there is no voltage drop due to the internal resistance when it is connected in series to measure the current.

<h3>What is voltmeter?</h3>

A voltmeter is a device, that can be used to measure the electric potential difference generated between the terminals of an electric circuit in volts.

An ideal voltmeter would have infinite resistance, because to ensure that there is no current is drawn by the voltmeter, when it is connected in parallel to measure the voltage.

Thus, we can conclude that, an ideal ammeter would have zero resistance, because to ensure that, there is no voltage drop due to the internal resistance. Similarly, an ideal voltmeter would have infinite resistance, because to ensure that there is no current is drawn by the voltmeter.

Learn more about the ammeter and voltmeter here:

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6 0

2 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

3 years ago