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4 years ago

11

A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the

electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.

1 answer:

Anton [14]4 years ago

6 0

Answer:

$1.176\times 10^{-3} N$

Upward

Explanation:

We are given that

Mass of scarp paper, $m=120 mg=120\times 10^{-6}kg$

1mg= $10^{-6} mkg$

Distance,d =8 mm= $8\times 10^{-3} m$

Magnitude of electric force = $F_E= w=mg$

Where $g=9.8 m/s^2$

Substitute the values

$F_E=120\times 10^{-6}\times 9.8$

$F_E=1.176\times 10^{-3} N$

Gravitational force act in downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

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s2008m [1.1K]

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

8 0

3 years ago

A boat moves through the water with two forces acting on it. One is a 1,800-N forward push by the water on the propeller, and th

baherus [9]

(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.

(c)Velocity at the end of that time will be 8.5 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

$\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f = 85 \ m$

c)

The velocity at the end of that time is found as;

$\rm v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5$ m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

#SPJ1

8 0

2 years ago

Which of the following best describes the function of a plant's flowers?

noname [10]

Answer: The function of a plant's flower is reproduce. Since the flowers are the reproductive organs of plant, they mediate the joining of the sperm, contained within pollen, to the ovules contained in the ovary.

Explanation:

8 0

3 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

3 years ago

Pc13 is a compound used to make pesticides. A reaction requires that 96.7 kg of pc13 be raised from 31.7 degrees celcius to 69.2

siniylev [52]

Answer:

Energy required = 3169.34 Joules.

Explanation:

The quantity of energy (Q) required can be determined by;

Q = mcΔθ

Where: m is the mass, c is the specific heat and Δθ is the change in temperature.

But, m = 96.7 kg, c = 0.874 J/(kg $^{o} C$ ), $T_{2}$ = $69.2^{o}$ and $T_{1}$ = $31.7^{o}$ .

So that,

Q = mc( $T_{2}$ - $T_{1}$ )

= 96.7 x 0.874 x ( $69.2^{o}$ - $31.7^{o}$ )

= 96.7 x 0.874 x 37.5

= 3169.3425

Q = 3169.34

= 3.2 KJ

The amount of energy required is 3169.34 Joules.

8 0

3 years ago