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3 years ago

The position of a charge in an electric field gives it _____electric

Physics

1 answer:

kykrilka [37]3 years ago

5 0

Answer:

The space surrounding a charged object is affected by the presence of the charge; an electric field is established in that space. A charged object creates an electric field - an alteration of the space or field in the region that surrounds it. Other charges in that field would feel the unusual alteration of the space.

Explanation:

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Which term describes what a british thermal unit measures?.

Yakvenalex [24]

Answer:

it is a measure of the heat content of fuels or energy sources

Explanation:

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2 years ago

What parts make up an atom

hram777 [196]

Answer:

Even though many super-tiny atomic particles exist, you only need to remember the three basic parts of an atom: electrons, protons, and neutrons. What are electrons, protons, and neutrons? Electrons are the smallest of the three particles that make up atoms.

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6 0

3 years ago

Read 2 more answers

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

anzhelika [568]

To solve the problem, use Kepler's 3rd law :

T² = 4π²r³ / GM

Solved for r :

r = [GMT² / 4π²]⅓

but first covert 6.00 years to seconds :

6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s

The radius of the orbit then is :

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m

6 0

3 years ago

A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.3

mixas84 [53]

Answer:

a) -2.34 m

b) -1.3 m/s

c) 0.056 m

d) 0.320 m/s

Explanation:

part a

Given:

s(0) = 0.27 m

v(0) = 0.14 m/s

a = -0.320 m/s^2

t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s(0) + v(0)*t + 0.5*a*t^2

s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

s(4.5) = -2.34 m

part b

Using kinematic equation of motion for constant acceleration:

v(t) = v(0) + a*t

v(4.5s) = (0.14) + (-0.32)(4.5) = -1.3 m/s

part c

Use equation for simple harmonic motion:

s(t) = A*cos(w*t)

v(t) = -A*w*sin (w*t)

a(t) = -A*(w)^2 * cos (w*t)

0.27 m = A*cos(w*t) .... Eq 1

0.14 m/s = - A*w*sin (w*t) .....Eq2

-0.320 = -A*(w)^2 * cos (w*t) .... Eq3

Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan(wt) = 0.14 / 0.27

tan(1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s(t) = 0.3*cos(1.0887*t)

s(4.5) = 0.3*cos(1.0887*4.5) = 0.056 m

part d

v(t) = - 0.32661 * sin (1.0887*t)

v(4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s

3 0

3 years ago

A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a  distance d. For the next loading?the spring is

Mrac [35]

Answer:

$W'=\dfrac{W}{9}$

Explanation:

The potential energy of the spring or the work done by the spring is given by :

$W=\dfrac{1}{2}kd^2$ ............(1)

k is the spring constant

d is the compression

When the spring is compressed a distance d' = d/3, let W' is the work is required to load the second dart. Then the work done is given by :

$W'=\dfrac{1}{2}kd'^2$

$W'=\dfrac{1}{2}k(d/3)^2$ .............(2)

Dividing equation (1) and (2) :

$\dfrac{W}{W'}=\dfrac{1/2kd^2}{1/2k(d/3)^2}$

$\dfrac{W}{W'}=9$

$W'=\dfrac{W}{9}$

So, the work required to load the second dart compared to that required to load the first is one-Ninth as much. Therefore, the correct option is (E).

7 0

3 years ago