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What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0

3 years ago

Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave

mylen [45]

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

7 0

3 years ago

A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina

elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car $a=1.6m/sec^2$ in negative direction so acceleration will be $a=-1.6m/sec^2$

From first equation of motion we know that

v = u+at

So $v=0+(-1.6)\times 5.8=-9.28m/sec$

So final velocity will be -9.28 m/sec

8 0

3 years ago

A tow truck is pulling 25,000 kg van with a tow hook that meets the van at an angle of 30° with the ground. How much force must

Law Incorporation [45]

Answer:

86605.08 N

Explanation:

The equation to calculate the force is:

Force = mass * acceleration

The force and the acceleration does not have the same direction in this case, so we need to decompose the force into its horizontal component, which is the force that will generate the horizontal acceleration:

Force_x = Force * cos(30)

Then, we have that:

Force_x = mass * acceleration

Force * cos(30) = 25000 * 3

Force * 0.866 = 75000

Force = 75000 / 0.866 = 86605.08 N

3 0

4 years ago

two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire

valentinak56 [21]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>

6 0

3 years ago