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3 years ago

15

NEED HELP ASAP!!!

1 answer:

TEA [102]3 years ago

3 0

Explanation:

Chem has to do with stored energy while nucleic deals with nucleic reactions.

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You're average speed would be 6km/hour or 100m per minute.

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True or false <br> A ball thrown horizontally.Gravity causes the ball to continue horizontally.

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3 years ago

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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th

GREYUIT [131]

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

$qvB=m\frac{v^2}{r}$

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

$v=\frac{qBr}{m}$

where in this problem we have:

$q=1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$B=208 G=208\cdot 10^{-4}T$ is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$

6 0

3 years ago

A bicycle travels 6.10 km due east in 0.210 h, then 11.30 km at 15.0° east of north in 0.560 h, and finally another 6.10 km due

Virty [35]

Answer:

Explanation:

Given

First bicycle travels 6.10 km due to east in 0.21 h

Suppose its position vector is $r_1$

$r_1=6.10\hat{i}$

After that it travels 11.30 km at $15^{\circ}$ east of north in 0.560 h

suppose its position vector is $r_2$

$r_{2}=11.30\left ( cos15\hat{j}+sin15\hat{i}\right )$

after that he finally travel 6.10 km due to east in 0.21 h

suppose its position vector is $r_3$

$r_{3}=6.10\hat{i}$

so position of final position is given by

$r=r_1+r_{2}+r_{3}$

$\vec{r}=15.12\hat{i}+10.91\hat{j}$

$\vec{v_{avg}}=\frac{\vec{r}}{t}$

t=0.21+0.56+0.21=0.98 h

$\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}$

$|v_{avg}|=\sqrt{361.71}=19.01 km/hr$

For direction

$tan\theta =\frac{11.13}{15.42}=0.721$

$\theta =35.791^{\circ}$ w.r.t to x axis

5 0

4 years ago