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3 years ago

The picture shows three sets of balloons, all with a particular charge. Which of the picture(s) is true? Please explain.

Physics

1 answer:

devlian [24]3 years ago

6 0

<span>Letter C has the correct illustration. Two objects with the same charge (in this case, both are positively charged) will repel each other. </span>

Letter A is incorrect because a positive charged object will attract a negatively charged object.

Letter B is incorrect because both of them are negatively charged, which means they should be repelling each other.

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A mass weighing 11 lb stretches a spring 4in. The mass is pulled down an additional 3 in and is then set in motion with an initi

Shtirlitz [24]

Answer:

a) $x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$ , b) $T = 3.628\,s$ , $A = 2.845\,ft$ , $\phi = -0.473\pi$

Explanation:

a) The system mass-spring is well described by the following equation of equilibrium:

$\Sigma F = k\cdot x - m\cdot g = m\cdot a$

After some handling in physical and mathematical definition, the following non-homogeneous second-order linear differential equation:

$\frac{d^{2}x}{dt^{2}}+\frac{k}{m}\cdot x = g$

The solution of this equation is:

$x (t) = A\cdot \cos \left(\sqrt{\frac{k}{m} } \cdot t + \phi\right)$

The velocity function is:

$v(t) = \sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi \right)$

Initial conditions are:

$x(0\,s) = 0.25\,ft, v(0\,s) = -5\,\frac{ft}{s}$

Equations at $t = 0\,s$ are:

$0.25\,ft = A\cdot \cos \phi\\-5\,\frac{ft}{s} =\sqrt{\frac{k}{m} }\cdot A\cdot \sin \phi$

The spring constant is:

$k = \frac{11\,lbf}{0.333\,ft}$

$k = 33\,\frac{lbf}{ft}$

After some algebraic handling, amplitude and phase angle are found:

$\phi = -0.473\pi$

$A = 2.845\,ft$

The position can be described by this function:

$x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$

b) The period of the motion is:

$T = \frac{2\pi}{\sqrt{\frac{k}{m} } }$

$T = 3.628\,s$

The amplitude is:

$A = 2.845\,ft$

The phase of the motion is:

$\phi = -0.473\pi$

8 0

4 years ago

A new moon is discovered orbiting Neptune with an orbital speed of 9.3 x103 m/s. Neptune's mass is 1.0 x1026 kg. A) What is the

AysviL [449]

The radius of the new moons orbit is R= 7.715 x 10^7 m
The orbital period of the moon is T= 14.48 hr

6 0

2 years ago

How much time would it take for an

Elina [12.6K]

36 and a half hours

Explanation:

4 0

4 years ago

A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe

Rasek [7]

Answer:

$56.5\ \text{s}$

$21.13\ \text{m/s}$

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}$

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is $12.5+39+5=56.5\ \text{s}$

Distance traveled

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}$

$s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}$

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2$

$s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}$

The total displacement of the car is $156.25+975+62.5=1193.75\ \text{m}$

Average velocity is given by

$\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}$

The average velocity of the car is $21.13\ \text{m/s}$ .

6 0

3 years ago

If the 78.0 kg astronaut were in a spacecraft 6R from the center of the earth, what would the astronaut's weight be on earth? 76

den301095 [7]

(a) 764.4 N

The weight of the astronaut on Earth is given by:

$F=mg$

where

m is the astronaut's mass

g is the acceleration due to gravity

Here we have

m = 78.0 kg

g = 9.8 m/s^2 at the Earth's surface

So the weight of the astronaut is

$F=(78.0)(9.8)=764.4 N$

(b) 21.1 N

The spacecraft is located at a distance of

$r=6R$

from the center of Earth.

The acceleration due to gravity at a generic distance r from the Earth's center is

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass.

We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:

$GM=9.8R^2$ (1)

the acceleration due to gravity at r=6R instead will be

$g'=\frac{GM}{(6R)^2}$

And substituting (1) into this formula,

$g'=\frac{9.8R^2}{36R^2}=0.27 m/s^2$

So the weight of the astronaut at the spacecratf location is

$F'=mg'=(78.0 kg)(0.27 m/s^2)=21.1 N$

6 0

3 years ago