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3 years ago

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A hollow sphere of radius 1.66 m is in a region where the electric field is radial and directed toward the center of the sphere.

If the magnitude of the field at the surface of the sphere is 21.5 N/C, what is the net electric flux through the spherical surface?

1 answer:

ladessa [460]3 years ago

5 0

Answer:

The value of flux will be "744.1 N.m²/C".

Explanation:

The given values are:

Magnitude,

E = 21.5 N/C

Radius,

R = 1.66 m

As we know,

⇒ $Flux = Area\times E$

On putting the estimated values, we get

⇒ $=-21.5\tines (4\times \pi\times 1.66^2 )$

⇒ $=744.1 \ N.m^2/C$

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A person is riding a bicycle, and its wheels have an angular velocity of 13.1 rad/s. Then, the brakes are applied and the bike i

lana [24]

First, let's convert the displacement to radians:

$10.9\text{ rev}=10.9\cdot2\pi\text{ rad}=21.8\pi\text{ rad}$

Now, let's calculate the angular acceleration using Torricelli's equation:

$\begin{gathered} V^2=V_o^2+2\cdot a\cdot d\\ \\ 0=13.1^2+2\cdot a\cdot21.8\pi\\ \\ 43.6\pi a=-171.61\\ \\ a=\frac{-171.61}{43.6\pi}\\ \\ a=-1.253\text{ rad/s^^b2} \end{gathered}$

Now, to calculate the time, we can use the formula below:

$\begin{gathered} V=V_o+at\\ \\ 0=13.1-1.253t\\ \\ 1.253t=13.1\\ \\ t=\frac{13.1}{1.253}\\ \\ t=10.45\text{ s} \end{gathered}$

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1 year ago

At what speed must a 0.6 kg stone be thrown in order that it has a relativistic mass of 0.76 kg? (c = 3.00 x 10^8 m/s) (A) 0.39

exis [7]

Explanation:

It is given that,

Relativistic Mass of the stone, m₀ = 0.6

Mass, $m=0.76\ kg$

Relativistic mass is given by :

$m=\dfrac{m_o}{\sqrt{1-\dfrac{v^2}{c^2}}}$ .........(1)

Where

c is the speed of light

On rearranging equation (1) we get :

$v^2=c^2(1-(\dfrac{m_o}{m})^2)$

$v=c\sqrt{ (1-(\dfrac{m_o}{m})^2)}$

$v=c\sqrt{ (1-(\dfrac{0.6}{0.76})^2)}$

v = 0.61378 c

or

v = 0.6138 c

So, the correct option is (c). Hence, this is the required solution.

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3 years ago

Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a

Tasya [4]

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

<h3>What is force?</h3>

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length, $\rm L_f = 1.45 m = 1450 \ mm$

Initial length, $\rm L_i = 7.5 mm$

$\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N$

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

To learn more about the force refer to the link;

brainly.com/question/26115859

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2 years ago

answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in

Natasha2012 [34]

Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

3 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago