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Maurinko [17]
3 years ago
11

A charged particle accelerates as it moves from location A to location B. If VA = 260 V and VB = 210 V, what is the sign of the

charged particle? positive negative (b) A electron loses electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?
Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

(a) Positive

(b) Electron gains energy as it moves from A to B.

Explanation:

VA = 260 V

VB = 210 V

An electron moves from lower to higher potential which is negatively charged and a positively charged particle moves  from higher to lower potential, so the charge particle is positive in nature.

(a) Positive

(b) No, electron gains energy as it moves from A to B.

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when you stretch a spring 13 cm past it's natural length, it excerts a force of 21 N. What is the spring constant of this spring
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So an alternate solution is

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An objects velocity will not change unless it is acted upon by ( balanced or unbalanced) forces?
OlgaM077 [116]
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Read 2 more answers
What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicl
Lapatulllka [165]

Answer:

276.74\times 10^8Mg/m^3

31.29 m/sec

Explanation:

We have given density of substance 0.14lb/in^3

We have convert this into Mg/m^3

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram =63.49\times 10^3\ Mg

We know that 1 in^3=1.6387\times 10^{-5}m^3

So 0.14in^3=0.14\times 1.6387\times 10^{-5}=0.2294\times 10^{-5}m^3

So 0.14lb/in^3 =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr =\frac{70\times 1609.34meter}{3600sec}=31.29m/sec

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3 years ago
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