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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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A dog pulls on a leash with a force of 15 23 is a big dog. The leash makes an angle of 36.7 degrees to the horizontal. What are

ch4aika [34]

Answer:

$x = 12.027N$ and $y = 8.964N$

Explanation:

The first sentence of this question is not explanatory enough. However, I'll assume the force to be 15N

$Force = 15N$

$\theta = 36.7$ to the horizontal

Required

Solve for the x and y components

Since the given angle is to the horizontal, the x and y coordinates are calculated using the following illustrations.

$Sin\theta = \frac{y}{Force}$ ---- y component

$Cos\theta = \frac{x}{Force}$ ---- x component

Calculating the y component.

Substitute 15 for Force and 36.7 for $\theta$

$Sin\theta = \frac{y}{Force}$ becomes

$Sin(36.7) = \frac{y}{15}$

Make y the subject

$y = 15 * Sin(36.7)$

$y = 15 * 0.5976$

$y = 8.964N$

Calculating the x component.

Substitute 15 for Force and 36.7 for $\theta$

$Cos\theta = \frac{x}{Force}$ becomes

$Cos(36.7) = \frac{x}{15}$

Make y the subject

$x = 15 * Cos(36.7)$

$x = 15 * 0.8018$

$x = 12.027N$

<em>Hence, the x and y components of the force are: 8.964N and 12.027N respectively.</em>

6 0

3 years ago

The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Flauer [41]

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

5 0

3 years ago

Moving current has electrical energy.

Inessa [10]

Yes, that’s true it has electrical energy

5 0

3 years ago

the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i

Naily [24]

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P = 0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

<u>P = 0.816 Watt</u>

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3 years ago

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Crank

Answer:

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8 0

3 years ago

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