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3 years ago

ElloNY1 ZINDA HERE xDWHAT IS MOLE ??

Physics

1 answer:

Julli [10]3 years ago

8 0

Answer:

yes I am

how are you

have a great day

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A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children pl

nordsb [41]

Answer:

Please find the complete solution in the attached file.

Explanation:

3 0

3 years ago

projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor

SIZIF [17.4K]

Answer:

Range, $R = MV²/2QE$

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

5 0

4 years ago

Someone help me! you get 27 points if you get it right

pashok25 [27]

Answer:

1. D

2. D

3. A

The reason why your body goes right and the car car goes left is because your body tries to stay where it was, which is on the right. Your upper body doesn't feel a force and because of that continues in the same direction. Your lower half is pulled out from under you by seat friction to the left, leaving you leaning to the right.

8 0

3 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time t (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time t (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the velocity at t = 8.00 s. Get the speed by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago

it is determined that a certain light wave has a wavelength of 3.012x10^-12 m. the light travels at 2.99x10^8 m/s. what is the f

Dahasolnce [82]

Answer:

$9.93\times 10^{19}Hz$

Explanation:

Speed of light is the product of its wavelength and frequency, expressed as

S=fw

Where s represent speed, f is frequency while w is wavelength

Making f the subject of the formula then

f=s/w

Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then

$f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz$

Therefore, the frequency equals to $9.93\times 10^{19}Hz$

4 0

3 years ago

ElloNY1 ZINDA HERE xDWHAT IS MOLE ??​

ElloNY1 ZINDA HERE xDWHAT IS MOLE ??