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navik [9.2K]
3 years ago
10

A 0.13 g plastic bead is charged by the addition of 1.0Ã1010 excess electrons. what electric field eâ (strength) will cause the

bead to hang suspended in the air?
Physics
1 answer:
slavikrds [6]3 years ago
5 0
The working equation for this is: E = F/Q, where E is the strength of electric field, F is the force and Q is the charge. The force is equal to:

F = mg = (0.13/1000 kg)(9.81 m/s²) = 1.2753×10⁻³ N

The charge of he excess electrons is equal to:
Q = (-1.6021766208×10⁻¹⁹ C/electron)(1×10¹⁰ electrons)
Q = -1.6021766208×10⁻⁹ C

E = 1.2753×10⁻³ N/-1.6021766208×10⁻⁹ C
E = -795,979.66 N/C
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Types of media that can influence one's social development include all of the following except
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A. movies b. internet c. books
6 0
3 years ago
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To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

8 0
3 years ago
21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest
zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

mgh=\frac{1}{2}mu^2

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}

where \Delta p is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

\Delta t=10.0 ms =0.010 s is the duration of the collision

Substituting,

F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0
3 years ago
a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at th
Elza [17]
... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m

(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry

... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
4 0
3 years ago
LOTS OF BRAINLIST WILL BE GIVING TO THOSE WHO HELP
Alex_Xolod [135]
You know you can skip those and just submit them, they don’t even check them
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