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3 years ago

8

how do i write this to where it is true? in the blue is a statement and the attached piece is false so i have to make it true an

d im struggling

1 answer:

OlgaM077 [116]3 years ago

4 0

Answer:

cross out the false piece in blue and write the true piece in red

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A box of groceries requires 5.0 newtons of force to lift it up 1.0 meter. How much work is done?

tigry1 [53]

Work is defined as the amount of force done multiplied by the distance it was applied. Since it is already given that the object of interest (a box of groceries) was lifted up 1 meter by a force of 5 newtons, work done can be solved using the following formula:

Work = Force * distance
Work = 5 N * 1 meter
Work = 5 N*m = 5 J

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4 years ago

If the frequency of the wave is 100 hz, what is the period of the wave?.

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Explanation:

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3 years ago

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3 years ago

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A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.

Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected is $B = 0.0036T$

Explanation:

From the question we are told that

The charge on the particle is $q = +2e$

The mass of the particle is $m = 6.64 *10^{-27} kg$

The potential difference is $V_a = 1.8 kV = 1.8 *10^{3} V$

The potential difference between the two parallel plate is $V_b = 120 V$

The separation between the plate is $d = 8 mm = \frac{8}{1000} = 8*10^{-3}m$

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV

$KE_b = PE_b$

Generelly

$KE_b = \frac{1}{2} m v^2$

And $PE_b = q V_a$

Equating this two formulas

$\frac{1}{2} mv^2 = q V_a$

making v the subject

$v = \sqrt{\frac{q V_a}{2 m} }$

Substituting value

$v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }$

$v = 41.65*10^4 m/s$

Generally the electric field between the plates is mathematically represented as

$E = \frac{V_b}{d}$

Substituting value

$E = \frac{120}{8*10^{-3}}$

$E = 15 *10^3 NC^{-1}$

the magnetic field is mathematically evaluate

$B = \frac{E}{v}$

$B = \frac{15 *10^{3}}{41.65 *10^4}$

$B = 0.0036T$

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3 years ago

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valentina_108 [34]

The answer is B- a positive charge

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