Elements from which two groups in the periodic table would most likely combine

An electron moves north at a velocity of 4.5 × 104 m/s and has a magnetic force of 7.2 × 10-18 n exerted on it. if the magnetic

Mice21 [21]

The magnetic force on a moving charge is given by:

$F=qvB \sin \theta$

where q is the charge, v is the speed of the charge, B is the magnetic field intensity and $\theta$ is the angle between the directions of B and v.

In our problem, the charge is an electron ( $q=1.6 \cdot 10^{-19}C$ ), the velocity of the charge is $v=4.5 \cdot 10^4 m/s$ , the magnetic force is $F=7.2 \cdot 10^{-18} N$ and the angle between the direction of v and B is $\theta=90^{\circ}$ , so $\sin \theta=1$ and we can ignore the sine in the formula. Therefore, if we rearrange the equation and we put these numbers in, we can find the intensity of the magnetic field:

$B=\frac{F}{qv}=\frac{7.2 \cdot 10^{-18} N}{(1.6 \cdot 10^{-19}C)(4.5 \cdot 10^{-4} m/s)}=0.001 T$

8 0

4 years ago

an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o

gladu [14]

The output of the machine is

(output work) = (output force) x (distance)

450 N-m = (output force) x (3 meters)

Divide each side
by 3 meters: Output force = (450 N-m) / (3 m)

= 150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m). So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.

5 0

3 years ago

An airplane flies 700 miles with a tail wind in 2.0 hrs. If it takes 2.5 hrs to cover the same distance against the headwind, th

nikklg [1K]

Answer:

B).315mph

Explanation:

Let the speed of the plane = p

Let the speed of the wind = w

Set up the system equation as;

Relative V: Time: Distance:

in wind direction: p + w 2 700

against wind: p - w 2.5 700

2(p + w) = 700

2.5(p - w) = 700

2p + 2w = 700

2.5p - 2.5w = 700

2.5 x: 5p + 5w = 1750

2 x: 5p - 5w = 1400

10p = 3150

p = 315 mph

Therefore, he speed of the plane in still air is 315 mph

6 0

4 years ago

several months after a forest is cut down, a nearby river becomes much muddier than ever before. Infer how the river changed

ryzh [129]

Well this is just my opinion but I think because all the plants were gone they cant stop mud from rolling into it. Also when it rains the trees wont suck up the water, that will make it over flow and release into the land or dirt on the outsides of it. Mixing those two gives mud and it can potentially flow back in the river in largeish masses causing the river to mix with the mud makinging it more muddy. Hope this helps :D

4 0

4 years ago

A projectile is fired straight upward with an initial veloc- ity of 100 m=s from the top of a building 20 m high and falls to th

SVETLANKA909090 [29]

Answer:

The answer to your question is:

a) h max = 529.7 m

b) t = 20.4 s

c) t = 20.6 s

Explanation:

a) h max = -(vo)² / 2g

= 100² / 2(9.81)

= 10000 / 19.62

= 509.7 m

total height = 509.7 + 20 = 529.7 m

b)

h = gt² / 2

t = √ 2h / g

t = √ 2(509.7)/9.81

t = √ 103.91

t = 10.19 s

total time = 2 x t = 2 x 10.19 = 20.4 s

c)

h = vot + 1/2gt²

20 = 100t + 1/2(9.91) t²

4.9t² + 100 t -20 = 0 quadratic equation

t = 0.19 s

Total time = 0.19 + 20.4 = 20.6 s

5 0

3 years ago