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3 years ago

NEED THIS ANSWERED ASAP!!!! IN MIDDLE OF TEST RN HELP PLZ

2 answers:

4 0

The answer should be A, it does not exhibit projectile motion and follows a straight path down the ramp.

It's not in the air or anything so we definitely know it's not a projectile :D.

Marina86 [1]3 years ago

3 0

The correct answer for this question would be option A. When Barry is conducting an experiment and rolls a tennis ball down a ramp, the statement that best describes the motion of the tennis ball is that, i<span>t does not exhibit projectile motion and follows a straight path down the ramp. Hope this helps.</span>

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

3 years ago

Read 2 more answers

How to calculate the uncertainty

agasfer [191]

Answer:

If you're taking the power of a number with an uncertainty, you multiply the relative uncertainty by the number in the power.

Explanation:

5 0

3 years ago

Which of the following is NOT a type of sediment?

MatroZZZ [7]

Answer:

It's soil.

Explanation:

7 0

2 years ago

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo

Advocard [28]

Answer:

laser's wavelength λ = 597.4 nm

Explanation:

Given:

Slit spacing, d = 1.17mm

Tenth bright fringe y = 4.57cm

Distance from slits, D = 8.95m

n = 10

λ = (d * y) / (D * n)

λ = (1.17x10⁻³ * 4.57x10⁻²) / (8.95 x 10)

λ = 5.3469‬x10⁻⁵ / 8.95x10¹

λ = 0.5974 x 10⁻⁵⁻¹

λ = 0.5974 x 10⁻⁶ m

λ = 597.4 x 10⁻⁹ m

λ = 597.4 nm

6 0

3 years ago