Can sound travel through a tree

A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t

svp [43]

Answer:

Explanation:

Given

for $\theta=10^{\circ}$

Sphere are $d=40\ cm$

when sphere $d_2=10\ cm$ apart suppose deflection is $\theta _2$

We know

$F=k_t\cdot \theta$

Where F=force between charged particle

$\theta =$ Deflection

$F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta$

$\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1$

thus $\theta \propto \frac{1}{r^2}$

for $\theta _2$

$\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2$

$\theta _2=16\times \theta _1$

$\theta _2=160^{\circ}$

(b)for $10^{\circ}$ deflection Potential $v_1=8\ kV$

Electric Potential is $V=\frac{kQ}{r}$

$Q=\frac{V\cdot r}{k}$

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

$\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}$

$\theta =\frac{V^2r^2}{k\cdot k_t}$

thus $\theta \propto V^2$

therefore

$\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2$

$\frac{10}{\theta _2}=(\frac{8}{4})^2$

$\theta _2=\frac{10}{4}=2.5^{\circ}$

5 0

2 years ago

1 What do the characteristics of electromagnetic radiation include?

statuscvo [17]

Answer:

first one is c second one is a 3 one is d 4 one is b

6 0

2 years ago

Calculate the acceleration of a car (in m/s^2) that accelerates from 0 to 30 m/s in 6 s along a straight road.

tester [92]

Answer: 5 m/s^2

Explanation: In order to solve this question we have to use the kinematic equation given by:

Vf= Vo+a*t where V0 is zero.

we know that it takes Vf( 30 m/s) in 6 seconds

so

a=(30 m/s)/6 s= 5 m/s^2

8 0

2 years ago

A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi

iVinArrow [24]

Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

mass of the small sphere, m = 2.30 Kg

when the Point is outside the sphere the Force between them is equal to

$F = \dfrac{GMm}{r^2}$ when r>R

When Point is inside the Sphere

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$ when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

$F = \dfrac{GMm}{r^2}$ [/tex]

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}$

F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}$

$F = 5.68\times 10^{-9}\ N$

3 0

2 years ago

An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. The car travels to the bottom, through a loop, an

aalyn [17]

Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.

Potential energy = (M) (G) (H) = (800) (9.8) (45) = 352,800 joules

It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.

Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules

So the coaster starts out with (352,000 + 6,400) = 359,200 joules of energy.

There's no friction, so it'll have that same energy at every point of the story.
=================================

Skip the loop for a moment, because the first question concerns the hill after
the loop. We'll come back to it.

The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is

(1/2) (M) (V)² = (400) (10)² = 40,000 joules.

Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.

PE = (M) (G) (H)

319,200 = (800) (9.8) (H)

H = (319,200) / (800 x 9.8) = 40.71 meters
=================================

Now back to the loop:

You said that the loop is 22m high at the top. The PE up there is

PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules

So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.

KE = (1/2) (M) (V)² = 186,720

(400) (V)² = 186,720

V² = 186,720 / 400 = 466.8

V = √466.8 = 21.61 m/s
===============================

Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.

At 4m off the ground, PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.

The rest will be kinetic. KE = (359,200 - 31,360) = 327,840 joules

KE = (1/2) (M) (V)² = 327,840

400 V² = 327,840

V² = 327,840 / 400 = 819.6

V = √819.6 = 28.63 m/s at the end
=======================================

If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .

If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere. Sorry.

6 0

2 years ago