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3 years ago

QUESTION 5

Physics

1 answer:

Fiesta28 [93]3 years ago

8 0

Answer:

D) sudden and irresistible on sets of sleep during normal waking periods

Explanation:

D because Narcolepsy (sudden and irresistible on sets of sleep during normal waking periods) is a part of insomnia

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(An easy problem which will be graded). Later in the quarter we will spend some time solving the diffusion equation Op(r, t) = D

alexandr402 [8]

Answer:

Explanation:

Answer is in the attachment below:

7 0

3 years ago

A rock falls off a cliff. How fast is the rock traveling vertically two seconds later?

Vikentia [17]

The rock it traveling really, really fast.
It is hard to exactly determine how fast bc u need the height of the cliff and how big the rock is.
Hope this helps and can I get brainliest answer!

8 0

3 years ago

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on

svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.

5 0

3 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

3 years ago

HELP ASAP!!! The gravitational pull of the Moon is much less than the

katovenus [111]

Answer:

I'm pretty sure its B and C

Explanation:

B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon

C bc mass is the same - you can't change it

7 0

3 years ago

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