Ask question

Ask question

All categories

3 years ago

5

A wall clock has a second hand 22.0 cmcm long. For related problem-solving tips and strategies, you may want to view a Video Tut

or Solution of Fast car, flat curve. Part A What is the radial acceleration of the tip of this hand

1 answer:

Alexus [3.1K]3 years ago

8 0

Answer:

Explanation:

The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression

ω²R where ω is angular velocity and R is radius of the circular path .

angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.

ω = 2π / T

= 2π / 60

= .1047

angular acceleration = .1047² x .22

= 2.41 x 10⁻³ rad / s² .

You might be interested in

When a tuning fork vibrates over an open pipe and the air in the pipe starts to vibrate, the vibrations in the tube are caused b

defon

Answer:

correct option is A. Resonance

Explanation:

solution

vibrations in the tube are caused by the Resonance because Resonance cause of sounds production in the musical instrument

As if resonance ( hollow cylindrical tube) immerse in cylinder of water and force in to vibration by tuning fork

As tines of tuning fork vibrate at natural frequency and it create sound wave that impinged up on opening of resonance tube

so correct option is A. Resonance

8 0

3 years ago

A thin, uniformly charged spherical shell has a potential of 832 V on its surface. Outside the sphere, at a radial distance of 2

Zolol [24]

Answer:

a

The radius is $r_1 = 0.315m$

b

The total charge is $Q= 2.912*10^{-8}C$

c

The electric potential inside sphere is the same with that outside the sphere which implies that the electric potential is 832 V

d

The magnitude of the electric field is $E= 2641.3 V/m$

e

The velocity is $v= 1.76 *10^{14} m/s$

Explanation:

From the question we are told that

The potential is $V_1 = 832 V$

The radial distance from the sphere is $d = 21.0cm = \frac{21}{100} = 0.21m$

The potential at the radial distance is $V_2 = 499V$

The potential at the surface of the sphere is mathematically represented as

$V = \frac{kQ}{r}$

$Vr = kQ$

Where kQ is a constant what this means that the the charge Q and the coulomb constant do not change

This means that

$V_1 r_1 = V_2 r_2$

Where $r_1$ is the radius of the sphere

and $r_2$ is the distance from that point where the second potential was measured to the center of the sphere which is mathematically represented as

$r_2 = r_1 + d$

Substituting this into the equation

$v_1 r_1 = V_2 (r_1 +d)$

Now substituting value

$832 * r_1 = 499 * (r_1 + 0.21)$

$832r_1 - 499r_1 = 104.79$

$333r_1 = 104.79$

$r_1 = \frac{104.79}{333}$

$r_1 = 0.315m$

From the equation above

$V = \frac{kQ}{r_1}$

making Q the subject

$Q = \frac{V r_1 }{k}$

k has a values of $k = 9*10^9 \ kg\cdot m^3 \cdot s^{-4} \cdot A^{-2}$

Substituting into the equation

$Q =\frac{832 * 0.315}{9*10^9}$

$Q= 2.912*10^{-8}C$

According to Gauss law the electric field from outside a sphere is taken to be an electric field from a point charge this mean that the potential outside a sphere is also taken as electric potential inside a sphere

The magnitude of a electric field from a sphere (point charge ) is mathematically represented as

$E = \frac{kQ}{r_1^2}$

Substituting values

$E = \frac{9*10^{9} * 2.912*10^{-8}}{0.315^2}$

$E= 2641.3 V/m$

According the the law of energy conservation

The electric potential energy at the point outside the surface where the second potential was measured(21 cm from the sphere surface) = The electric potential energy at the surface + The kinetic energy of the charge (electron) at that the surface

Generally Electric potential energy is mathematically represented as

$EPE = V * e$

Where is e is an electron

And Kinetic energy is mathematically represented as

$KE = \frac{1}{2} m v^2$

From the statement above

$V_2 e = V_1 e + \frac{mv^2}{2}$

But from the question we can deduce that the potential at the surface is zero

So the equation becomes

$V_2 e = \frac{mv^2}{2}$

The charge an electron has a value $e = 1.602*10^{-19}C$

And the mass of an electron is $m = 9.109 *10^{-31}kg$

Making v the subject

$v = \sqrt{\frac{2V_2 e}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 499 * 1.602 *10^{-19}}{9.109*10^{-31}} }$

$v= 1.76 *10^{14} m/s$

7 0

3 years ago

A soccer ball was kicked . it had the mass of .42 kg and accelerated at 25m/s . what was the force?

kaheart [24]

brainly.com/question/3414828
here lol

6 0

3 years ago

If you go to the beach on a hot summer day, the temperature of the sand is much higher than the temperature of the water. If we

guapka [62]

Water has a higher specific heat so it requires more energy to heat than sand.

7 0

3 years ago

Read 2 more answers

Scientists in the laboratory create a uniform electric field e⃗ = 6. 0×105 k^v/m in a region of space where b⃗ =0⃗

zvonat [6]

Scientist in the laboratory create uniform electric field.

PART A: The electric field in rocket frame is (0,0, 6.0 × 10 ⁵).

PART B: The magnetic field in the rocket frame is 7.3 × 10⁻⁶ j∧ .

<h3>What is uniform electric field?</h3>

The electric field whose strength is not changing with respect to time is called uniform magnetic field.

The electric field is 6.0 × 10 ⁵ k^v/m and the magnetic field is zero.

E' = E + (V×B)

= 6.0 × 10 ⁵ + (1.1 × 10⁶ × 0)

E' = 6.0 × 10 ⁵ k∧

(Ex , Ey, Ez) = (0,0, 6.0 × 10 ⁵)

The new magnetic field is represented as

B' = B - 1 /c²(v × E)

=0 - 1 / (3× 10⁸)²(1.1 × 10⁶ i∧ × 6.0 × 10 ⁵k∧ )

B' = 7.3 × 10⁻⁶ j∧

PART A: The electric field in rocket frame is (0,0, 6.0 × 10 ⁵).

PART B: The magnetic field in the rocket frame is 7.3 × 10⁻⁶ j∧ .

Learn more about uniform electric field.

brainly.com/question/26446532

#SPJ4

7 0

2 years ago