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Aliun [14]
1 year ago
7

A charged particle decelerates as it moves from location a to location b. if va = 160 v and vb = 100 v, what is the sign of the

charged particle?
Physics
1 answer:
skelet666 [1.2K]1 year ago
3 0

The sign of the charged particle is positively charged.

<h3>What is potential difference?</h3>
  • When a single charge is transported in an electric field, work is done by the potential difference (also known as electrical potential).
  • There is potential energy stored in this charge that could flow when work is done on it.
  • Voltage is the possibility of a single charge flowing. The need to flow increases with voltage.
  • Here, voltage can be the potential differences.

The potential difference between the 2 points determines the movement of that particle. An electron moves from lower to higher potential which is negatively charged, and a positively charged particle moves  from higher to lower potential.

Now, since the particle is moving from a point A having 160 v potential to point B having 100 v potential that is it is moving from higher potential to a lower potential therefore the particle will be a positively charged one.

Learn more about potential difference,

brainly.com/question/23716417

#SPJ1

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Earth's surface receives about twice as much energy from the atmosphere than from the sun as a result of ____.
Oxana [17]
I think the answer is “greenhouse effect”
5 0
3 years ago
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?
iren2701 [21]

Answer:

  • 4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

V(r) = \frac{4}{3} \ \pi \ r^3

If we got an uncertainty \Delta r the formula for the uncertainty of V is:

\Delta V(r) = \sqrt{  (\frac{dV}{dr} \Delta r)^2  }

We can calculate this uncertainty, first we obtain the derivative:

\frac{dV}{dr}  = 3 * \frac{4}{3} \ \pi \ r^2

\frac{dV}{dr}  = 4 \ \pi \ r^2

And using it in the formula:

\Delta V(r) = \sqrt{  (4 \ \pi \ r^2\Delta r)^2  }

\Delta V(r) = \sqrt{  4^2 \ \pi^2 \ r^4 \Delta r^2  }

\Delta V(r) =  4 \  \pi \ r^2 \Delta r

The relative uncertainty is:

\frac{\Delta V(r)}{V(r)}

\frac{ 4 \  \pi \ r^2 \Delta r  }{ \frac{4}{3} \ \pi \ r^3}

\frac{ 3  \Delta r  }{  r}

Using the values for the problem:

\frac{ 3 * 0.09 m  }{  5.66 m} = 0.0477

This is, a percent uncertainty of 4.77 %

4 0
3 years ago
What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
3 years ago
A 0.0100-kilogram grape hangs from a vine 1.50 meters from the ground. What is the potential energy of the grape?
ludmilkaskok [199]
The potential energy is defined as Ep=m*g*h where m is the mass of the body, g=9.81 m/s² and h is the height of the body. In our case m=0.01 kg and h=1.5 m. So when we input the values into the equation:

Ep=0.01*9.81*1.5= 0.14715 J. 

So the potential energy of a grape is Ep=0.14715 J. 
8 0
3 years ago
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