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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

1 answer:

Svetlanka [38]3 years ago

6 0

The formula written in the 3rd line above the picture is WRONG. Don't use it. Use the formula the way it's printed in the picture.

V = d / t

That means Speed = (distance) / (time)

The question tells us that v = 330 m/s

So you write 330 m/s in the equation in place of 'v', like this:

330 m/s = (distance) / (time)

The question also tells us that the time is 0.4 second

So you write 0.4 sec in place of 'time', like this:

330 m/s = (distance) / (0.4 second)

Finally, you take this, and multiply each side of the equation by (0.4 sec). Then it'll say

distance = (330 m/s) x (0.4 second)

As soon as you do that one single multiplication there with your pencil or your calculator, you'll have the distance.

This is either the 2nd or 3rd time you've posted this same exact question since last weekend. It can be solved THIS time exactly like the answers that were posted those other times.

The DOT in the picture is marked for the wrong choice. Use the formula that's printed in the picture, not copied above it.

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In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro

blondinia [14]

In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass.
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were.

Ok, Rameses:

Elementary charge . . . . . 1.6 x 10⁻¹⁹ coulomb
negative on the electron
plussitive on the proton

Electron rest-mass . . . . . 9.11 x 10⁻³¹ kg

a). The force between two charges is

F = (9 x 10⁹) Q₁ Q₂ / R²

= (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

      = ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

=    8.05 x 10⁻⁸ Newton .

b). Centripetal acceleration =

   v² / r .

A = (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

   =    7.7 x 10²² m/s² .

That's an enormous acceleration ... about 7.85 x 10²¹ G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use " F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether F = M A .

I'm going to leave that step for you.   Good luck !

4 0

3 years ago

A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning

zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

$L_{i}=L_{f}$

Here, = the system's angular momentum before the collision

$L_{i}$ = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here, = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

4 0

1 year ago

Read 2 more answers

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di

S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F $\alpha$ e

F=ke

k=f/e.........*

where k is the force constant or the constant of proportionality

k=f/e

$f_{eff} =f_{1} +f_{2}$ ............................1

also for effective force constant

divide all through by extension

1) Total force is

Ft=F1+F2

Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

Substituting e1 and e2, you get

1/keq=1/k1+1/k2

Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.

4 0

3 years ago

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Mademuasel [1]

Here it is. *WARNING* VERY LONG ANSWER

________________________________________...
11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) 

The change in PE =mgh=5*9.8*12=588 J 
______________________________________... 
12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. 

Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, 

Their KE as they crossed the line=(1/2)Mv^2 

Their KE as they crossed the line=0.5*550*(15.98)^2 

Their KE as they crossed the line is 70224.11 J 

______________________________________... 
13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m 

She trips and drops the spare tire of mass = m = 10.0 kg, 

The tire rolls down the hill with an intial speed = u = 2.00 m/s. 

The height of top of the next hill = h = 5.00 m 

Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 

Initial total mechanical energy =mgH+(1/2)mu^2 

Suppose the final speed at the top of second hill is v 

Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 

As mechanical energy is conserved, 

Final total mechanical energy =Initial total mechanical energy 

mgh+(1/2)mv^2=mgH+(1/2)mu^2 

v = sq rt [u^2+2g(H-h)] 

v = sq rt [4+2*9.8(20-5)] 

v = sq rt 298 

v =17.2627 m/s 

The speed of the tire at the top of the next hill is 17.2627 m/s 
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14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. 

a.)The mass of bean = m = 2.0 g 

Height up to which the been jumps = h = 1.0 cm from hand 

Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg 

b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s 
_____________________________ 
15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. 

The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s 
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16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, 

The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 
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EDIT 
1.) A train is accelerating at a rate = a = 2.0 km/hr/s. 

Acceleration 

Initial velocity = u = 20 km/hr, 

Velocity after 30 seconds = v = u + at 

Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = 

Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr 

Velocity after 30 seconds = v = 80 km/hr=22.22 m/s 
_______________________________- 
2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. 

His acceleration = a =11.1/9=1.233 m/s^2 

Distance he covered = s = (1/2)at^2=49.95 m

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