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tamaranim1 [39]

3 years ago

10

Really easy it's physics.

2 answers:

iragen [17]3 years ago

4 0

I think it’s B because we aren’t looking for Vf

joja [24]3 years ago

4 0

B is the correct answer

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8) Studies have shown that wider highway lanes lead to fewer accidents. Which factor most likely prevents states from widening r

Bess [88]

Probably D, though I may be wrong

8 0

4 years ago

Read 2 more answers

A rectangular loop consists of 80 turns, each carrying a current 8

liberstina [14]

Magnetic moment of loop is given by the formula

$M = NiA$

$M = 80*8 *0.4*0.3$

$M = 76.8$

now magnetic field is given here

$B = 40 * 10^{-3} T$

Now the torque on the loop is given by

$T = MBsin\theta$

$T = 76.8* 40*10^{-3}*sin30$

$T = 1.536 Nm$

so above is the torque on the loop.

3 0

4 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

4 years ago

What is charge measured in

Gre4nikov [31]

Charge is measured in coulombs.

6 0

4 years ago

Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Aleonysh [2.5K]

De broglie wavelength, $\lambda = \frac{h}{mv}$ , where h is the Planck's constant, m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom, $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0

3 years ago