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3 years ago

What is a point value for an arrow that is on two different colors?

1 answer:

8 0

It depends on what it is closest to but I would say for instance black is 5 points and red is 6 if u land on the line 5.5

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A charged particle moves through a magnetic field. In which situation is the magnetic force zero?

maksim [4K]

Answer:

The answer is the option a.

Explanation:

We know that magnetic force (Fm) is defined as

Fm = q (v x B)

Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.

"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is

v x B = vBsin (thetha)

So,

Fm = qvBsin (thetha)

And, in case in which v and B are parallel vectors, thetha is zero, and,

sin (thetha)=sin (0) = 0

So, Fm=0

7 0

3 years ago

Express 50,000 m/s2 in scientific notation. 5.0 x 103m/s2 0.5 x 105m/s2 5.0 x 104m/s2

Tresset [83]

Answer would be 5.0x104m/s2

8 0

3 years ago

Read 2 more answers

Two people are pushing on a desk to the right with a force 15n each what is the net force on the desk

dybincka [34]

30N to the right direction

3 0

3 years ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago

What happens to the free energy released as electrons are passed from photosystem II to photosystem I through a series of electr

lana [24]

<span>It is used to establish and maintain a proton gradient.</span>

6 0

3 years ago