What should you do if you need to pull over to get some rest?

Which effect has been useful (and successful) in the search for and identification of black holes in the universe

kvasek [131]

Answer:

Detection of x-rays from a binary star undergoing mass exchange, where mass of component star can be determined.

4 0

2 years ago

In the illustration of the solar spectrum, the upper left portion of the spectrum shows the __________ visible light.

Usimov [2.4K]

Lowest frequency visible light

8 0

3 years ago

A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

5 0

3 years ago

A roller coaster starts at the top of a hill of height h, goes down the hill, and does a circular loop of radius r before contin

jeka94

a) See free-body diagram in attachment

b) Net force in the y-direction: $F_y=mg+N$ [/tex]

c) The velocity at which the roller coaster will fall is $[tex]v=\sqrt{gr}$ [/tex]

d) The speed of the roller coaster must be 17.1 m/s

e) The roller coaster should start from a height of 90 m

f) The roller coaster should start from a height of 100 m

Explanation:

a)

See the free-body diagram in attachment. There are only two forces acting on the roller coaster at the top of the loop:

The weight of the roller coaster, acting downward, indicated by $mg$ (where m is the mass of the roller coaster and g is the acceleration of gravity)
The normal reaction exerted by the track on the roller coaster, acting downward, and indicated with N

The two forces are represented in the diagram as two downward arrows (the length is not proportional to their magnitude, in this case)

b)

Since there are only two forces acting on the roller coaster at the top of the loop, and both forces are acting downward, then we can write the vertical net force as follows (we take downward as positive direction):

$F_y = mg + N$

where

mg is the weight

N is the normal reaction

Since the roller coaster is in circular motion, this net force must be equal to the centripetal force, therefore

$m\frac{v^2}{r}=mg+N$

where v is the speed of the car at the top of the loop and r is the radius of the loop.

c)

For this part of the problem, we start from the equation written in part b)

$m\frac{v^2}{r}=mg+N$

where the term on the left represents the centripetal force, and the terms on the right are the weight and the normal reaction.

We now re-arrange the equation making v (the speed) as the subject:

$v=\sqrt{gr+\frac{Nr}{m}}$

However, the velocity at which the roller coaster will fall is the velocity at which the normal reaction becomes zero (the roller coaster loses contact with the track), so when

N = 0

And as a result, the minimum velocity of the cart is

$v=\sqrt{gr}$

d)

In this part, we are told that the radius of the loop is

r = 30 m

And the mass of the cart is

m = 50 kg

Moreover, the acceleration of gravity is

$g=9.8 m/s^2$

We said that the minimum velocity that the cart must have in order not to fall at the top is

$v=\sqrt{gr}$

And substituting, we find

$v=\sqrt{(9.8)(30)}=17.1 m/s$

e)

According to the law of conservation of energy, the initial gravitational energy of the roller coaster at the starting point must be equal to the sum of the kinetic energy + gravitational potential energy at the top of the loop, therefore:

$mgh = \frac{1}{2}mv^2 + mg(2r)$