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Svetach [21]
2 years ago
12

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

he same nail in a ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes a 13.0 ∘ angle with the vertical. How many surplus electrons are on each sphere?
Physics
1 answer:
Musya8 [376]2 years ago
3 0

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0
3 years ago
A body weighing 250 grams was dropped from a helicopter flying at an altitude of 100 meters. Determine the potential energy of t
kogti [31]

Answer:

the potential energy of this body is 245 J.

Explanation:

Given;

mass of the body, m = 250 g = 0.25 kg

height from which the body was dropped, h = 100 m

acceleration due to gravity, g = 9.8 m/s²

The potential energy of this body is calculated as;

P.E = mgh

substitute the given values and solve for the potential energy of this body;

P.E = 0.25 x 9.8 x 100

P.E = 245 J.

Therefore, the potential energy of this body is 245 J.

7 0
2 years ago
HELP HELP HELP ME!!!
zzz [600]
I believe the answer is C. Hope this helps!!

7 0
3 years ago
If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?
weqwewe [10]

Answer:

Time, t = 13.34 seconds.

Explanation:

Given the following data;

Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s

Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s

Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²

To find the time;

Acceleration = (v - u)/t

-0.833 = (12.5 - 23.61)/t

-0.833t = -11.11

t = 11.11/0.833

Time, t = 13.34 seconds.

6 0
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The path the moon's umbra traces across earth's surface is the path of totality.What would you see if you were standing in the p
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You would see the darkest part of the moon
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