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2 years ago

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Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} E ⃗ =(4000 j ^ +3000 k ^ ) N/

C. What is its electric flux through a circular area of radius 1.83 m that lies in the xy-plane?

1 answer:

likoan [24]2 years ago

4 0

Answer:

Electric flux $\phi=31562.63\ Nm^2/C$

Explanation:

Given that,

Electric field acting on the circular area, $E=(4000j+3000k)\ N/C$

We need to find the electric flux through a circular area of radius 1.83 m that lies in the xy-plane. It lies in xy plane, such that the area vector point in z direction. The electric flux is given by :

$\phi=E{\cdot}A$

$\phi=(4000j+3000k){\cdot}Ak$

Using dot product properties, we get the value of electric flux as :

$\phi=3000\times Ak$

$\phi=3000\times \pi (1.83)^2$

$\phi=31562.63\ Nm^2/C$

So, the electric flux through a circular area is $\phi=31562.63\ Nm^2/C$ . Hence, this is the required solution.

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A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

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<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity $V_{esc}$ is given by the following equation:

$V_{esc}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M$ is the mass of the asteroid

$R$ is the radius of the asteroid

On the other hand, we know the density of the asteroid is $\rho=3.84(10)^{8}g/m^{3}$ and its volume is $V=2.17(10)^{12}m^{3}$ .

The density of a body is given by:

$\rho=\frac{M}{V}$ (2)

Finding $M$ :

$M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})$ (3)

$M=8.33(10)^{20}g=8.33(10)^{17}kg$ (4) This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

$V=\frac{4}{3}\piR^{3}$ (5)

Finding $R$ :

$R=\sqrt[3]{\frac{3V}{4\pi}}$ (6)

$R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}$ (7)

$R=8031.38m$ (8) This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

$V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}$ (9)

Finally:

$V_{esc}=117.626m/s$ This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

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2 years ago

Which of the following represents a convex lens?<br> A. +f<br> B. -di<br> C. -f<br> D. +di

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F is the focal length of the lens. A positive f indicates a converging optical device, like a convex lens or a concave mirror.

Choice A

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A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the en

Nina [5.8K]

Answer:

B) d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

Em₀ = U = m g h₁

Final point. Lower (slide bottom)

Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

mgh₁ = ½ m v² + mgh₂

v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

x = v₀ₓ t

y = t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

t = √ 2h₂ / g

We substitute in the other equation

d = √ (2g (h₁-h₂)) √ 2h₂ / g

d = √ (4 h₂ (h₁-h₂))

H = h₁ + h₂

h₁ = H -h₂

d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

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2 years ago

When jumping, a flea accelerates at an astounding 1000 m/s2 but over the very short distance of 0.50 mm. If a flea jumps straigh

Nadusha1986 [10]

Answer:

The flea reaches a height of 51 mm.

Explanation:

Hi there!

The equations of height and velocity of the flea are the following:

During the jump:

h = h0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

While in free fall:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the flea at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration of the flea due to the jump.

v = velocity of the flea at time t.

g = acceleration due to gravity.

First, let's calculate how much time it takes the flea to reach a height of 0.0005 m. With that time, we can calculate the speed reached by the flea during the jump:

h = h0 + v0 · t + 1/2 · a · t²

If we place the origin of the frame of reference on the ground, then, h0 = 0. Since the flea is initially at rest, v0 = 0. Then:

h = 1/2 · a · t²

We have to find the value of t for which h = 0.0005 m:

0.0005 m = 1/2 · 1000 m/s² · t²

0.0005 m / 500 m/s² = t²

t = 0.001 s

Now, let's find the velocity reached in that time:

v = v0 + a · t (v0 = 0)

v = a · t

v = 1000 m/s² · 0.001 s

v = 1.00 m/s

When the flea is at a height of 0.50 mm, its velocity is 1.00 m/s. This initial velocity will start to decrease due to the downward acceleration of gravity. When the velocity is zero, the flea will be at the maximum height. Using the equation of velocity, let's find the time at which the flea is at the maximum height (v = 0):

v = v0 + g · t

At the maximum height, v = 0:

0 m/s = 1.00 m/s - 9.81 m/s² · t

-1.00 m/s / -9.81 m/s² = t

t = 0.102 s

Now, let's find the height reached by the flea in that time:

h = h0 + v0 · t + 1/2 · g · t²

h = 0.0005 m + 1.00 m/s · 0.102 s - 1/2 · 9.81 m/s² · (0.102 s)²

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The flea reaches a height of 51 mm.

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2 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

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