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3 years ago

Help! 50 points each and brainliest!!!

Physics

2 answers:

klio [65]3 years ago

7 0

Answer:

Your answer is the following...

(b)"Many valence electrons are shared between the atoms"

Explanation:

inna [77]3 years ago

3 0

Answer:

Explanation:

edge 2020

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Aerogenerators are used to produce electricity from

ki77a [65]

Like windmills they use the winds to generate their power.

4 0

3 years ago

An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

kupik [55]

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

4 0

4 years ago

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

Yanka [14]

Answer:

The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}$

$E=2.41\times10^{3}\ V/m$

Hence, The electric field due to the right ring at a location midway between the two rings is $2.41\times10^{3}\ V/m$

3 0

4 years ago

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s

yarga [219]

Answer:

$I=336.6kgm/s$

Explanation:

The equation for the linear impulse is as follows:

$I=F\Delta t$

where $I$ is impulse, $F$ is the force, and $\Delta t$ is the change in time.

The force, according to Newton's second law:

$F=ma$

and since $a=\frac{v_{f}-v_{i}}{\Delta t}$

the force will be:

$F=m(\frac{v_{f}-v_{i}}{\Delta t})$

replacing in the equation for impulse:

$I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)$

we see that $\Delta t$ is canceled, so

$I=m(v_{f}-v_{i})$

And according to the problem $v_{i}=0m/s$ , $v_{f}=5.10m/s$ and the mass of the passenger is $m=66kg$ . Thus:

$I=(66kg)(5.10m/s-0m/s)$

$I=(66kg)(5.10m/s)$

$I=336.6kgm/s$

the magnitude of the linear impulse experienced the passenger is $336.6kgm/s$

6 0

3 years ago

Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal

NeX [460]

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

8 0

3 years ago