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RUDIKE [14]
3 years ago
5

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

l to the ground, and a vertical uniform electric field of strength 260N/C passes upward through the tetrahedron.
What is the electric flux through the base?







What is the electric flux through each of the three sides?
Physics
1 answer:
Jet001 [13]3 years ago
6 0

Answer:

-7.03645640575 Nm²/C

2.34548546858 Nm²/C

Explanation:

S = Surface area

Electric flux through the base is given by

\phi_b=-\int Eds\\\Rightarrow \phi_b=-ES\\\Rightarrow \phi_b=-260\dfrac{\sqrt{3}}{4}0.25^2\\\Rightarrow \phi_b=-7.03645640575\ Nm^2/C

The electric flux through the base is -7.03645640575 Nm²/C

Inside the tetrahedron there is no charge so total flux will be zero

\phi_b+3\phi_s=0\\\Rightarrow \phi_s=-\dfrac{\phi_b}{3}\\\Rightarrow \phi_s=-\dfrac{-7.03645640575}{3}\\\Rightarrow \phi_s=2.34548546858\ Nm^2/C

The electric flux through the sides is 2.34548546858 Nm²/C

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8TH GRADE SCIENCE I NEED NOW DO NOT SKIP
yan [13]

Explanation:

1. Force=mass*acceleration

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What is the momentum of a vehicle that has a mass of 800 kg and is moving at a velocity of 5 m/s
yarga [219]

Answer:

4000 kg.m/s

Explanation:

p=mv

m=800 kg

v=5 m/s

p=(800)(5)= 4000 kg.m/s

3 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0
3 years ago
Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is tw
seropon [69]

Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

F=ma\\\Rightarrow a_l=\frac{F}{m_l}

F=ma\\\Rightarrow a_j=\frac{F}{2m_l}\\\Rightarrow a_j=\frac{1}{2}a_l

v=u+at\\\Rightarrow v_l=0+\frac{F}{m_l}\times t\\\Rightarrow v_l=a_lt

v=u+at\\\Rightarrow v_l=0+\frac{F}{2m_l}\times t\\\Rightarrow v_j=\frac{1}{2}a_lt\\\Rightarrow v_j=\frac{1}{2}v_l\\\Rightarrow v_l=2v_j

Hence, Lilly's speed is two times John's speed.

4 0
3 years ago
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