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den301095 [7]
3 years ago
14

Two metal spheres are suspended from strings. The

Physics
1 answer:
IRINA_888 [86]3 years ago
6 0

Answer:

The answer to your question is the letter B. A repulsive force of 8.0 x 10³N

Explanation:

Data

Charge 1 = -2.0 x 10² C

Charge 2 = -4.0 x 10⁻⁸ C

K (Coulomb constant) = 9 x 10⁹ Nm²/C²

distance = 3 m

Formula

F = K (q₁ x q₂) / r²

Substitution

F = 9 x 10⁹ (-2.0 x 10²)(-4.0 x 10⁻⁸) / 3²

Simplification

F = 9 x 10⁹ (0.000008) / 9

F = 72000 / 9

Result

F = 8000 C = 8 x 10³ N

Conclusion

As both spheres are negative the force is negative with magnitude 8x 10³N

You might be interested in
An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-
maw [93]

Answer:

  1. Power requirement <u>P</u> for the banner is found to be  30.62 W
  2. Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
  3. Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The <em>Cd </em>for banner was given, whereas the <em>Cd </em>for a flat plate is generally found to be around <em><u>1.28</u></em><em> </em>which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
  4. Power requirement <u>P</u> for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

  1. v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
  2. The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
  3. The power can be determined this equation: P = F . v, where F represents <em>the drag-force</em> that we will need to determine and v represents the<em> velocity of the airplane</em>
  4. The equation to determine drag-force is: F = 1/2 * d *  C_d * A

In the drag-force equation Cd represents the c<em>o-efficient of drag</em> and A represents the <em>frontal area of the banner/plate/balloon (the object being towed)</em>

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

<u>Part a)</u> We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = <em>1/2 * 0.06* 1.225 * 20</em> = 0.735 N. Now to find the power P we will use P = F . v i.e.<em> 0.735 * 41.66</em> = <u><em>30.62 W</em></u>

<em></em>

<u>Part b) </u>For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = <em>1/2 * 1.28 * 1.225 * 20</em> = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = <u><em>653.225 W</em></u>

<u><em></em></u>

<u>Part c)</u> The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

<u>Part d)</u> The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : <em>pi* r^2 (r = d / 2).</em> Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = <u><em>40.08 W</em></u>

4 0
3 years ago
Which describes an object in projectile motion? Check all that apply.
Elodia [21]

Answer:

Gravity acts to pull the object down.

The object’s inertia carries it forward.

The path of the object is curved.

Explanation:

The motion of a projectile consists of two separate motions:

- A uniform motion along the horizontal direction, where the velocity is constant; since there are no forces along this direction, the velocity does not change, and so the object continues its motion for inertia --> so, the statement  "The object’s inertia carries it forward" is true.

- A uniformly accelerated motion along the vertical direction, with a constant downward acceleration (g=9.8 m/s^2, acceleration due to gravity). So, the vertical velocity changes, due to the presence of the gravity that acts to pull the object down.

- As a result of the combination of these two motions, the object follows a curved path (in particular, it is a parabolic path).

5 0
3 years ago
1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande
kvv77 [185]

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

5 0
3 years ago
1) Which of the following statements about physical models is not true?
Alex_Xolod [135]
1. "<span>Physical models are usually in the form of a graph" is the statement among the statements given about physical models that is not true. The correct option among all the options that are given in the question is the third option or option "c".

2. "Predicting weather patterns" is the one among the following scenarios that </span>represents the best use of a scientific model. <span>The correct option among all the options that are given in the question is the first option or option "a".</span>
8 0
3 years ago
What is the measurement of loudness of sounds
AURORKA [14]
The measurement of sound is in decibels.
4 0
3 years ago
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