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3 years ago

Compare and Contrast How is a tug - of - war similar to unequal sharing of electron

1 answer:

7 0

Imagine that the two atoms are playing tug of war with the electron, but one is pulling with a stronger force than the other. This inequality results in the majority of the rope being pulled toward the stronger atom. In the bond, the electron is pulled closer to the stronger atom too. This results in each side having a partial charge, one being more negative than the other.

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Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ

bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c, s= 400 m

r= ?

we have to solve for the radius

we know that circumference

$c= 2\pi r$

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular displcament

∅= 400/63.63

∅= 6.28 radians

8 0

3 years ago

A war wolf is a device used during the middle ages to assault fortifications with large rocks. A simple trebuchet is constructed

Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object $m_1=52 kg$

distance of $m_1$ from the axle $r_1=14 cm$

mass of rock $m_2=123 gm$

Length of rod $=4.1 m$

distance of $m_2$ from axle $r_2=4.1-0.14=3.96 m$

Net torque acting is

$T_{net}=m_1gr_1-m_2gr_2$

$T_{net}=52\times 0.14\times g-0.123\times 3.96\times g$

$T_{net}=6.793\times 9.8$

$T_{net}=66.57 N-m$

Work done by $T_{net}$ is converted to rock kinetic Energy

thus

$T_{net}\times \theta =\frac{mv^2}{2}$

Where $\theta =angle\ turned =\frac{\pi }{2}$

$v= velocity\ at\ launch$

$66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}$

$v^2=66.57\times \pi$

$v=\sqrt{1700.511}$

$v=41.23 m/s$

3 0

3 years ago

It is important that your muscles are very warm when doing this type of stretching.

MaRussiya [10]

Answer:

ballistic stretching

3 0

3 years ago

Read 2 more answers

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

The another equation of motion is $v^2=u^2+2as$ using this equation for second plane $40^2=0+2\times 2.67\times s$

s = 300 m

5 0

3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago